39:I[811522,["/_next/static/immutable/chunks/18pod8863va08.js","/_next/static/immutable/chunks/2x8o0dxuycygx.js","/_next/static/immutable/chunks/2tgra8qb3l1t-.js","/_next/static/immutable/chunks/2ofq0h3cxt38w.js","/_next/static/immutable/chunks/2zzmbopslaqoj.js","/_next/static/immutable/chunks/0vlnvmaep27te.js","/_next/static/immutable/chunks/1t-qox8vqf86a.js","/_next/static/immutable/chunks/2gu_fgkdkpxlx.js","/_next/static/immutable/chunks/3e9_duuz9q1nl.js","/_next/static/immutable/chunks/1xx_b342vgjad.js"],"SidebarProvider"] 3a:I[469067,["/_next/static/immutable/chunks/18pod8863va08.js","/_next/static/immutable/chunks/2x8o0dxuycygx.js","/_next/static/immutable/chunks/2tgra8qb3l1t-.js","/_next/static/immutable/chunks/2ofq0h3cxt38w.js","/_next/static/immutable/chunks/2zzmbopslaqoj.js","/_next/static/immutable/chunks/0vlnvmaep27te.js","/_next/static/immutable/chunks/1t-qox8vqf86a.js","/_next/static/immutable/chunks/2gu_fgkdkpxlx.js","/_next/static/immutable/chunks/3e9_duuz9q1nl.js","/_next/static/immutable/chunks/1xx_b342vgjad.js"],"ItemsProvider"] 3b:T43f,## The question How do you go deep without turning the work into an indefinitely growing pile of reading, branches, and experiments? ## Working model A useful research step must still be able to change a decision. Pin the exact source and the result owner, preserve the first surprising observation, and make each plausible explanation face a test that could make it lose. Separate “this mechanism can work” from “this is the smallest justified change.” Stop when the next probe cannot change the selected mechanism, claim boundary, or reopening condition. A careful negative result is a real result when it records what lost, what remains possible, and what new evidence would justify reopening. ## Why it matters Without a discriminator, more depth mostly creates narrative momentum. With one, even a long investigation keeps shrinking the decision surface. ## Provenance Synthesized from Linux Fieldwork at revision `c79d34b65fb2cd8c54234f361f073248c53b513a`. The source note describes a research method; it does not claim every investigation needs the same machinery.3c:T475,

The question

How do you go deep without turning the work into an indefinitely growing pile of reading, branches, and experiments?

Working model

A useful research step must still be able to change a decision. Pin the exact source and the result owner, preserve the first surprising observation, and make each plausible explanation face a test that could make it lose. Separate “this mechanism can work” from “this is the smallest justified change.”

Stop when the next probe cannot change the selected mechanism, claim boundary, or reopening condition. A careful negative result is a real result when it records what lost, what remains possible, and what new evidence would justify reopening.

Why it matters

Without a discriminator, more depth mostly creates narrative momentum. With one, even a long investigation keeps shrinking the decision surface.

Provenance

Synthesized from Linux Fieldwork at revision c79d34b65fb2cd8c54234f361f073248c53b513a. The source note describes a research method; it does not claim every investigation needs the same machinery.

3d:T40c,## The question When a multi-source install command backs up a destination, fails to copy one source, then sees another source with the same name, which state owns the path? ## Working model There are two nearby but independent machines. **Invocation ownership** asks whether an earlier operand completed its data-copy phase for this destination. Only a completed copy owns the name; a source-stat failure, no-op comparison, or failed copy does not. **Backup transaction** asks whether the old destination was moved aside and the replacement copy committed. If the backup exists but the copy fails, remove any partial destination and restore the backup. Failures after a completed copy do not roll the old file back. The useful boundary is the commit event—not the function's final success boolean. ## Provenance Synthesized from black-box GNU `install` behavior and controlled uutils work recorded at Linux Fieldwork revision `c79d34b65fb2cd8c54234f361f073248c53b513a`. It is a design model, not an upstream acceptance claim.3e:T468,

The question

When a multi-source install command backs up a destination, fails to copy one source, then sees another source with the same name, which state owns the path?

Working model

There are two nearby but independent machines.

Invocation ownership asks whether an earlier operand completed its data-copy phase for this destination. Only a completed copy owns the name; a source-stat failure, no-op comparison, or failed copy does not.

Backup transaction asks whether the old destination was moved aside and the replacement copy committed. If the backup exists but the copy fails, remove any partial destination and restore the backup. Failures after a completed copy do not roll the old file back.

The useful boundary is the commit event—not the function's final success boolean.

Provenance

Synthesized from black-box GNU install behavior and controlled uutils work recorded at Linux Fieldwork revision c79d34b65fb2cd8c54234f361f073248c53b513a. It is a design model, not an upstream acceptance claim.

3f:T40f,

The question

What happens if an HTTP library removes chunk framing from a body, but a proxy forwards Transfer-Encoding: chunked unchanged?

Working model

Chunked encoding describes wire bytes. A client library may consume chunk sizes, separators, the terminal chunk, and trailers before returning only payload bytes. If the proxy sends those decoded bytes while advertising chunked encoding, the next client tries to parse payload text as chunk sizes.

Headers must describe the representation actually sent on the next hop. Remove stale transfer framing and hop-by-hop fields, including fields named by Connection, then choose a valid downstream boundary: correct length, newly generated chunk framing, or close delimiting.

A correct cache file does not prove the first streamed response was valid.

Provenance

Synthesized from a loopback regression against Python's HTTP stack at Linux Fieldwork revision c79d34b65fb2cd8c54234f361f073248c53b513a.

40:T451,

The question

A shell wrapper can observe command failure, subordinate failure, a signal, and cleanup failure. Why does “last error wins” describe control flow rather than meaning?

Working model

Capture the primary result before cleanup and encode precedence explicitly. A primary command or first handled-signal failure outranks a subordinate status channel; that outranks the first cleanup failure; success comes last.

Do not share one $?-based trap body across EXIT, INT, and TERM. A signal may be deferred while the shell waits, so $? can describe the completed child instead of the signal. Give signal handlers explicit statuses, prevent EXIT from running cleanup twice, and stabilize the first handled result while cleanup continues.

Fallible result-file reads also need containment so set -e cannot manufacture a new final failure.

Provenance

Synthesized from Linux Fieldwork wrapper regressions at revision c79d34b65fb2cd8c54234f361f073248c53b513a.

41:T433,

The question

Why is forwarding dict(request.headers) from a proxy to an origin both a security bug and a compatibility bug?

Working model

Headers have scope. Proxy-Authorization is a credential for the proxy, not a server selected by the request URL. Connection controls apply to one transport hop. Connection can name additional hop-specific fields, and header names are case-insensitive.

Construct the origin request from an explicit policy: remove proxy credentials and hop-by-hop fields, parse every Connection token, preserve one valid Host, and retain safe end-to-end fields—including repeated fields when their meaning requires it.

Origin Authorization is different from Proxy-Authorization; deleting both is not a safe shortcut.

Provenance

Synthesized from a loopback mmdebstrap proxy regression at Linux Fieldwork revision c79d34b65fb2cd8c54234f361f073248c53b513a. It does not define every extension or CONNECT policy.

42:T2696,""" PROBLEM: Fraudulent Transaction Detection Stripe processes billions of transactions daily. To protect merchants, we need to identify potentially fraudulent transactions based on user behavior patterns. A transaction is considered SUSPICIOUS if ANY of the following conditions are met: 1. The user makes more than 3 transactions within any 60-second window 2. The user's transactions in any 60-second window exceed $3000 total 3. The user makes a transaction from a different city within 30 minutes of a previous transaction (impossible travel) You are given a list of transactions, where each transaction is formatted as: "timestamp,user_id,amount,city" - timestamp: Unix timestamp (integer, seconds) - user_id: String identifier for the user - amount: Transaction amount in dollars (float) - city: City where transaction occurred (string) Return a list of transaction indices (0-indexed) that are SUSPICIOUS, sorted in ascending order. EXAMPLES: Input: [ "1609459200,user1,100.00,NYC", # 0: OK "1609459210,user1,200.00,NYC", # 1: OK "1609459220,user1,300.00,NYC", # 2: OK "1609459230,user1,400.00,NYC", # 3: SUSPICIOUS (4th transaction in 60s) "1609459300,user2,3500.00,LA", # 4: SUSPICIOUS (exceeds $3000) "1609462000,user1,50.00,Chicago", # 5: OK (2800 seconds / ~46 min later) ] Output: [3, 4] Input: [ "1609459200,user1,100.00,NYC", # 0: OK "1609459500,user1,200.00,LA", # 1: SUSPICIOUS (different city within 30 min) ] Output: [1] CONSTRAINTS: - 1 <= len(transactions) <= 100,000 - All timestamps are valid Unix timestamps - 0 < amount <= 100,000 - user_id contains only alphanumeric characters - city contains only alphabetic characters """ from collections import defaultdict from typing import List def find_suspicious_transactions(transactions: List[str]) -> List[int]: """ Identify suspicious transactions based on fraud detection rules. Time Complexity: O(n * m) where n = number of transactions, m = max transactions per user in a 60-second window Space Complexity: O(n) for storing parsed transactions """ # Parse all transactions parsed = [] for i, txn in enumerate(transactions): parts = txn.split(',') timestamp = int(parts[0]) user_id = parts[1] amount = float(parts[2]) city = parts[3] parsed.append({ 'index': i, 'timestamp': timestamp, 'user_id': user_id, 'amount': amount, 'city': city }) # Group transactions by user for efficient processing user_transactions = defaultdict(list) for txn in parsed: user_transactions[txn['user_id']].append(txn) # Sort each user's transactions by timestamp for user_id in user_transactions: user_transactions[user_id].sort(key=lambda x: x['timestamp']) suspicious = set() for user_id, txns in user_transactions.items(): for i, current_txn in enumerate(txns): current_time = current_txn['timestamp'] # Check rules within 60-second window (Rules 1 & 2) window_start = current_time - 60 window_txns = [] window_amount = 0 for j in range(i, -1, -1): # Look backwards from current if txns[j]['timestamp'] >= window_start: window_txns.append(txns[j]) window_amount += txns[j]['amount'] else: break # Rule 1: More than 3 transactions in 60-second window if len(window_txns) > 3: suspicious.add(current_txn['index']) # Rule 2: Total amount exceeds $3000 in 60-second window if window_amount > 3000: suspicious.add(current_txn['index']) # Rule 3: Different city within 30 minutes (impossible travel) if i > 0: thirty_min_ago = current_time - 1800 # 30 * 60 = 1800 seconds for j in range(i - 1, -1, -1): prev_txn = txns[j] if prev_txn['timestamp'] >= thirty_min_ago: if prev_txn['city'] != current_txn['city']: suspicious.add(current_txn['index']) break else: break return sorted(list(suspicious)) # ============================================================================ # TEST CASES # ============================================================================ def test_find_suspicious_transactions(): # Test Case 1: More than 3 transactions in 60 seconds transactions1 = [ "1609459200,user1,100.00,NYC", "1609459210,user1,200.00,NYC", "1609459220,user1,300.00,NYC", "1609459230,user1,400.00,NYC", # 4th transaction - suspicious "1609459300,user2,100.00,LA", ] result1 = find_suspicious_transactions(transactions1) assert result1 == [3], f"Test 1 failed: expected [3], got {result1}" print("✓ Test 1 passed: Detects >3 transactions in 60s window") # Test Case 2: Amount exceeds $3000 in 60-second window transactions2 = [ "1609459200,user1,1000.00,NYC", "1609459210,user1,1500.00,NYC", "1609459220,user1,600.00,NYC", # Total now $3100 - suspicious ] result2 = find_suspicious_transactions(transactions2) assert result2 == [2], f"Test 2 failed: expected [2], got {result2}" print("✓ Test 2 passed: Detects amount >$3000 in 60s window") # Test Case 3: Impossible travel (different city within 30 min) transactions3 = [ "1609459200,user1,100.00,NYC", "1609459500,user1,200.00,LA", # Different city, only 5 min later ] result3 = find_suspicious_transactions(transactions3) assert result3 == [1], f"Test 3 failed: expected [1], got {result3}" print("✓ Test 3 passed: Detects impossible travel") # Test Case 4: Same city within 30 min is OK transactions4 = [ "1609459200,user1,100.00,NYC", "1609459500,user1,200.00,NYC", # Same city - OK ] result4 = find_suspicious_transactions(transactions4) assert result4 == [], f"Test 4 failed: expected [], got {result4}" print("✓ Test 4 passed: Same city transactions are not flagged") # Test Case 5: Multiple users don't affect each other transactions5 = [ "1609459200,user1,100.00,NYC", "1609459210,user2,200.00,LA", "1609459220,user1,300.00,NYC", "1609459230,user2,400.00,LA", ] result5 = find_suspicious_transactions(transactions5) assert result5 == [], f"Test 5 failed: expected [], got {result5}" print("✓ Test 5 passed: Different users are independent") # Test Case 6: Multiple rules violated transactions6 = [ "1609459200,user1,2000.00,NYC", "1609459210,user1,500.00,NYC", "1609459220,user1,300.00,NYC", "1609459230,user1,400.00,LA", # >3 txns, >$3000, different city ] result6 = find_suspicious_transactions(transactions6) assert result6 == [3], f"Test 6 failed: expected [3], got {result6}" print("✓ Test 6 passed: Multiple violations on same transaction") # Test Case 7: Window resets after 60 seconds transactions7 = [ "1609459200,user1,100.00,NYC", "1609459210,user1,200.00,NYC", "1609459220,user1,300.00,NYC", "1609459400,user1,400.00,NYC", # 200 seconds later - new window "1609459410,user1,500.00,NYC", "1609459420,user1,600.00,NYC", ] result7 = find_suspicious_transactions(transactions7) assert result7 == [], f"Test 7 failed: expected [], got {result7}" print("✓ Test 7 passed: 60-second window resets properly") # Test Case 8: Large single transaction transactions8 = [ "1609459200,user1,3500.00,NYC", # Single transaction >$3000 ] result8 = find_suspicious_transactions(transactions8) assert result8 == [0], f"Test 8 failed: expected [0], got {result8}" print("✓ Test 8 passed: Single large transaction flagged") # Test Case 9: Edge case - exactly 3 transactions (should be OK) transactions9 = [ "1609459200,user1,100.00,NYC", "1609459210,user1,200.00,NYC", "1609459220,user1,300.00,NYC", ] result9 = find_suspicious_transactions(transactions9) assert result9 == [], f"Test 9 failed: expected [], got {result9}" print("✓ Test 9 passed: Exactly 3 transactions is not suspicious") # Test Case 10: Edge case - exactly $3000 (should be OK) transactions10 = [ "1609459200,user1,1000.00,NYC", "1609459210,user1,1000.00,NYC", "1609459220,user1,1000.00,NYC", ] result10 = find_suspicious_transactions(transactions10) assert result10 == [], f"Test 10 failed: expected [], got {result10}" print("✓ Test 10 passed: Exactly $3000 is not suspicious") print("\n" + "="*50) print("All tests passed! ✓") print("="*50) if __name__ == "__main__": test_find_suspicious_transactions() # Example from problem description print("\n--- Running Example from Problem Description ---\n") example = [ "1609459200,user1,100.00,NYC", "1609459210,user1,200.00,NYC", "1609459220,user1,300.00,NYC", "1609459230,user1,400.00,NYC", "1609459300,user2,3500.00,LA", "1609462000,user1,50.00,Chicago", # 46 min later, different city OK ] result = find_suspicious_transactions(example) print(f"Input transactions: {len(example)} transactions") print(f"Suspicious indices: {result}") print(f"Expected: [3, 4]")43:T1303b,

"""
PROBLEM: Fraudulent Transaction Detection

Stripe processes billions of transactions daily. To protect merchants, we need to 
identify potentially fraudulent transactions based on user behavior patterns.

A transaction is considered SUSPICIOUS if ANY of the following conditions are met:
1. The user makes more than 3 transactions within any 60-second window
2. The user's transactions in any 60-second window exceed $3000 total
3. The user makes a transaction from a different city within 30 minutes of a previous 
   transaction (impossible travel)

You are given a list of transactions, where each transaction is formatted as:
    "timestamp,user_id,amount,city"
    
- timestamp: Unix timestamp (integer, seconds)
- user_id: String identifier for the user
- amount: Transaction amount in dollars (float)
- city: City where transaction occurred (string)

Return a list of transaction indices (0-indexed) that are SUSPICIOUS, sorted in 
ascending order.

EXAMPLES:

Input: [
    "1609459200,user1,100.00,NYC",      # 0: OK
    "1609459210,user1,200.00,NYC",      # 1: OK
    "1609459220,user1,300.00,NYC",      # 2: OK
    "1609459230,user1,400.00,NYC",      # 3: SUSPICIOUS (4th transaction in 60s)
    "1609459300,user2,3500.00,LA",      # 4: SUSPICIOUS (exceeds $3000)
    "1609462000,user1,50.00,Chicago",   # 5: OK (2800 seconds / ~46 min later)
]
Output: [3, 4]

Input: [
    "1609459200,user1,100.00,NYC",      # 0: OK
    "1609459500,user1,200.00,LA",       # 1: SUSPICIOUS (different city within 30 min)
]
Output: [1]

CONSTRAINTS:
- 1 <= len(transactions) <= 100,000
- All timestamps are valid Unix timestamps
- 0 < amount <= 100,000
- user_id contains only alphanumeric characters
- city contains only alphabetic characters

"""

from collections import defaultdict
from typing import List


def find_suspicious_transactions(transactions: List[str]) -> List[int]:
    """
    Identify suspicious transactions based on fraud detection rules.
    
    Time Complexity: O(n * m) where n = number of transactions, 
                     m = max transactions per user in a 60-second window
    Space Complexity: O(n) for storing parsed transactions
    """
    
    # Parse all transactions
    parsed = []
    for i, txn in enumerate(transactions):
        parts = txn.split(',')
        timestamp = int(parts[0])
        user_id = parts[1]
        amount = float(parts[2])
        city = parts[3]
        parsed.append({
            'index': i,
            'timestamp': timestamp,
            'user_id': user_id,
            'amount': amount,
            'city': city
        })
    
    # Group transactions by user for efficient processing
    user_transactions = defaultdict(list)
    for txn in parsed:
        user_transactions[txn['user_id']].append(txn)
    
    # Sort each user's transactions by timestamp
    for user_id in user_transactions:
        user_transactions[user_id].sort(key=lambda x: x['timestamp'])
    
    suspicious = set()
    
    for user_id, txns in user_transactions.items():
        for i, current_txn in enumerate(txns):
            current_time = current_txn['timestamp']
            
            # Check rules within 60-second window (Rules 1 & 2)
            window_start = current_time - 60
            window_txns = []
            window_amount = 0
            
            for j in range(i, -1, -1):  # Look backwards from current
                if txns[j]['timestamp'] >= window_start:
                    window_txns.append(txns[j])
                    window_amount += txns[j]['amount']
                else:
                    break
            
            # Rule 1: More than 3 transactions in 60-second window
            if len(window_txns) > 3:
                suspicious.add(current_txn['index'])
            
            # Rule 2: Total amount exceeds $3000 in 60-second window
            if window_amount > 3000:
                suspicious.add(current_txn['index'])
            
            # Rule 3: Different city within 30 minutes (impossible travel)
            if i > 0:
                thirty_min_ago = current_time - 1800  # 30 * 60 = 1800 seconds
                
                for j in range(i - 1, -1, -1):
                    prev_txn = txns[j]
                    if prev_txn['timestamp'] >= thirty_min_ago:
                        if prev_txn['city'] != current_txn['city']:
                            suspicious.add(current_txn['index'])
                            break
                    else:
                        break
    
    return sorted(list(suspicious))


# ============================================================================
# TEST CASES
# ============================================================================

def test_find_suspicious_transactions():
    # Test Case 1: More than 3 transactions in 60 seconds
    transactions1 = [
        "1609459200,user1,100.00,NYC",
        "1609459210,user1,200.00,NYC",
        "1609459220,user1,300.00,NYC",
        "1609459230,user1,400.00,NYC",  # 4th transaction - suspicious
        "1609459300,user2,100.00,LA",
    ]
    result1 = find_suspicious_transactions(transactions1)
    assert result1 == [3], f"Test 1 failed: expected [3], got {result1}"
    print("✓ Test 1 passed: Detects >3 transactions in 60s window")
    
    # Test Case 2: Amount exceeds $3000 in 60-second window
    transactions2 = [
        "1609459200,user1,1000.00,NYC",
        "1609459210,user1,1500.00,NYC",
        "1609459220,user1,600.00,NYC",  # Total now $3100 - suspicious
    ]
    result2 = find_suspicious_transactions(transactions2)
    assert result2 == [2], f"Test 2 failed: expected [2], got {result2}"
    print("✓ Test 2 passed: Detects amount >$3000 in 60s window")
    
    # Test Case 3: Impossible travel (different city within 30 min)
    transactions3 = [
        "1609459200,user1,100.00,NYC",
        "1609459500,user1,200.00,LA",  # Different city, only 5 min later
    ]
    result3 = find_suspicious_transactions(transactions3)
    assert result3 == [1], f"Test 3 failed: expected [1], got {result3}"
    print("✓ Test 3 passed: Detects impossible travel")
    
    # Test Case 4: Same city within 30 min is OK
    transactions4 = [
        "1609459200,user1,100.00,NYC",
        "1609459500,user1,200.00,NYC",  # Same city - OK
    ]
    result4 = find_suspicious_transactions(transactions4)
    assert result4 == [], f"Test 4 failed: expected [], got {result4}"
    print("✓ Test 4 passed: Same city transactions are not flagged")
    
    # Test Case 5: Multiple users don't affect each other
    transactions5 = [
        "1609459200,user1,100.00,NYC",
        "1609459210,user2,200.00,LA",
        "1609459220,user1,300.00,NYC",
        "1609459230,user2,400.00,LA",
    ]
    result5 = find_suspicious_transactions(transactions5)
    assert result5 == [], f"Test 5 failed: expected [], got {result5}"
    print("✓ Test 5 passed: Different users are independent")
    
    # Test Case 6: Multiple rules violated
    transactions6 = [
        "1609459200,user1,2000.00,NYC",
        "1609459210,user1,500.00,NYC",
        "1609459220,user1,300.00,NYC",
        "1609459230,user1,400.00,LA",  # >3 txns, >$3000, different city
    ]
    result6 = find_suspicious_transactions(transactions6)
    assert result6 == [3], f"Test 6 failed: expected [3], got {result6}"
    print("✓ Test 6 passed: Multiple violations on same transaction")
    
    # Test Case 7: Window resets after 60 seconds
    transactions7 = [
        "1609459200,user1,100.00,NYC",
        "1609459210,user1,200.00,NYC",
        "1609459220,user1,300.00,NYC",
        "1609459400,user1,400.00,NYC",  # 200 seconds later - new window
        "1609459410,user1,500.00,NYC",
        "1609459420,user1,600.00,NYC",
    ]
    result7 = find_suspicious_transactions(transactions7)
    assert result7 == [], f"Test 7 failed: expected [], got {result7}"
    print("✓ Test 7 passed: 60-second window resets properly")
    
    # Test Case 8: Large single transaction
    transactions8 = [
        "1609459200,user1,3500.00,NYC",  # Single transaction >$3000
    ]
    result8 = find_suspicious_transactions(transactions8)
    assert result8 == [0], f"Test 8 failed: expected [0], got {result8}"
    print("✓ Test 8 passed: Single large transaction flagged")
    
    # Test Case 9: Edge case - exactly 3 transactions (should be OK)
    transactions9 = [
        "1609459200,user1,100.00,NYC",
        "1609459210,user1,200.00,NYC",
        "1609459220,user1,300.00,NYC",
    ]
    result9 = find_suspicious_transactions(transactions9)
    assert result9 == [], f"Test 9 failed: expected [], got {result9}"
    print("✓ Test 9 passed: Exactly 3 transactions is not suspicious")
    
    # Test Case 10: Edge case - exactly $3000 (should be OK)
    transactions10 = [
        "1609459200,user1,1000.00,NYC",
        "1609459210,user1,1000.00,NYC",
        "1609459220,user1,1000.00,NYC",
    ]
    result10 = find_suspicious_transactions(transactions10)
    assert result10 == [], f"Test 10 failed: expected [], got {result10}"
    print("✓ Test 10 passed: Exactly $3000 is not suspicious")
    
    print("\n" + "="*50)
    print("All tests passed! ✓")
    print("="*50)


if __name__ == "__main__":
    test_find_suspicious_transactions()
    
    # Example from problem description
    print("\n--- Running Example from Problem Description ---\n")
    example = [
        "1609459200,user1,100.00,NYC",
        "1609459210,user1,200.00,NYC",
        "1609459220,user1,300.00,NYC",
        "1609459230,user1,400.00,NYC",
        "1609459300,user2,3500.00,LA",
        "1609462000,user1,50.00,Chicago",  # 46 min later, different city OK
    ]
    result = find_suspicious_transactions(example)
    print(f"Input transactions: {len(example)} transactions")
    print(f"Suspicious indices: {result}")
    print(f"Expected: [3, 4]")

44:Tbcb,from typing import List, Dict, Tuple class LedgerSystem: def __init__(self): # Determine strictness of processing self.balances: Dict[str, int] = {} # To track duplicates: Set of (timestamp, user_id, amount) self.seen_transactions = set() def parse_transaction(self, transaction_str: str) -> Tuple[int, str, int]: """ Helper to cleanly parse the input string. Returns: (timestamp, user_id, amount) """ parts = transaction_str.split(',') timestamp = int(parts[0].strip()) user_id = parts[1].strip() amount = int(parts[2].strip()) return timestamp, user_id, amount def process_transactions(self, transactions: List[str]) -> Dict[str, int]: self.balances = {} self.seen_transactions = set() # Step 1: Parse all transactions first parsed_data = [] for t_str in transactions: parsed_data.append(self.parse_transaction(t_str)) # Step 2: Sort by timestamp (Required for Part 3) # Python's sort is stable and efficient (Timsort) parsed_data.sort(key=lambda x: x[0]) for timestamp, user, amount in parsed_data: # Part 2: Duplicate Check # We construct a unique signature for the transaction txn_signature = (timestamp, user, amount) if txn_signature in self.seen_transactions: continue self.seen_transactions.add(txn_signature) # Part 3: Overdraft Check current_balance = self.balances.get(user, 0) if current_balance + amount < 0: # Reject transaction (do nothing) # In a real interview, you might want to log this rejection continue # Apply transaction self.balances[user] = current_balance + amount return self.balances # --- Test Cases (How you should verify your code in the interview) --- def run_tests(): solver = LedgerSystem() # Test Part 1: Basic logic t1 = [ "100, user_a, 100", "105, user_b, 50", "110, user_a, -20" ] assert solver.process_transactions(t1) == {'user_a': 80, 'user_b': 50}, "Failed Part 1" # Test Part 2: Duplicates t2 = [ "100, user_a, 100", "100, user_a, 100", # Dup "110, user_a, 50" ] assert solver.process_transactions(t2) == {'user_a': 150}, "Failed Part 2" # Test Part 3: Chronological Order & Overdraft t3 = [ "200, user_a, -50", # Should process 2nd (Balance 100 -> 50) "100, user_a, 100", # Should process 1st (Balance 0 -> 100) "300, user_a, -60" # Should process 3rd (Balance 50 -> -10 REJECT) ] result_t3 = solver.process_transactions(t3) assert result_t3 == {'user_a': 50}, f"Failed Part 3. Got {result_t3}" print("All tests passed!") if __name__ == "__main__": run_tests()45:T7ef5,

from typing import List, Dict, Tuple

class LedgerSystem:
    def __init__(self):
        # Determine strictness of processing
        self.balances: Dict[str, int] = {}
        # To track duplicates: Set of (timestamp, user_id, amount)
        self.seen_transactions = set()

    def parse_transaction(self, transaction_str: str) -> Tuple[int, str, int]:
        """
        Helper to cleanly parse the input string.
        Returns: (timestamp, user_id, amount)
        """
        parts = transaction_str.split(',')
        timestamp = int(parts[0].strip())
        user_id = parts[1].strip()
        amount = int(parts[2].strip())
        return timestamp, user_id, amount

    def process_transactions(self, transactions: List[str]) -> Dict[str, int]:
        self.balances = {}
        self.seen_transactions = set()
        
        # Step 1: Parse all transactions first
        parsed_data = []
        for t_str in transactions:
            parsed_data.append(self.parse_transaction(t_str))
        
        # Step 2: Sort by timestamp (Required for Part 3)
        # Python's sort is stable and efficient (Timsort)
        parsed_data.sort(key=lambda x: x[0])
        
        for timestamp, user, amount in parsed_data:
            # Part 2: Duplicate Check
            # We construct a unique signature for the transaction
            txn_signature = (timestamp, user, amount)
            if txn_signature in self.seen_transactions:
                continue
            self.seen_transactions.add(txn_signature)

            # Part 3: Overdraft Check
            current_balance = self.balances.get(user, 0)
            
            if current_balance + amount < 0:
                # Reject transaction (do nothing)
                # In a real interview, you might want to log this rejection
                continue
            
            # Apply transaction
            self.balances[user] = current_balance + amount
            
        return self.balances

# --- Test Cases (How you should verify your code in the interview) ---

def run_tests():
    solver = LedgerSystem()
    
    # Test Part 1: Basic logic
    t1 = [
        "100, user_a, 100",
        "105, user_b, 50",
        "110, user_a, -20"
    ]
    assert solver.process_transactions(t1) == {'user_a': 80, 'user_b': 50}, "Failed Part 1"

    # Test Part 2: Duplicates
    t2 = [
        "100, user_a, 100",
        "100, user_a, 100", # Dup
        "110, user_a, 50"
    ]
    assert solver.process_transactions(t2) == {'user_a': 150}, "Failed Part 2"

    # Test Part 3: Chronological Order & Overdraft
    t3 = [
        "200, user_a, -50",  # Should process 2nd (Balance 100 -> 50)
        "100, user_a, 100",  # Should process 1st (Balance 0 -> 100)
        "300, user_a, -60"   # Should process 3rd (Balance 50 -> -10 REJECT)
    ]
    result_t3 = solver.process_transactions(t3)
    assert result_t3 == {'user_a': 50}, f"Failed Part 3. Got {result_t3}"

    print("All tests passed!")

if __name__ == "__main__":
    run_tests()

46:T4d5,class BrowserHistory: def __init__(self, homepage: str): # Initialize with the homepage self.history = [homepage] self.pos = 0 # Current index self.top = 0 # The current 'end' of valid history def visit(self, url: str) -> None: self.pos += 1 # If we are visiting a new page from the middle of history, # we overwrite the next slot. if self.pos < len(self.history): self.history[self.pos] = url else: self.history.append(url) # After a visit, the 'top' is always our current position # effectively deleting any forward history. self.top = self.pos def back(self, steps: int) -> str: # Move back, but don't go past index 0 self.pos = max(0, self.pos - steps) return self.history[self.pos] def forward(self, steps: int) -> str: # Move forward, but don't go past the 'top' self.pos = min(self.top, self.pos + steps) return self.history[self.pos] # Your BrowserHistory object will be instantiated and called as such: # obj = BrowserHistory(homepage) # obj.visit(url) # param_2 = obj.back(steps) # param_3 = obj.forward(steps)47:T43d1,

class BrowserHistory:
    def __init__(self, homepage: str):
        # Initialize with the homepage
        self.history = [homepage]
        self.pos = 0    # Current index
        self.top = 0    # The current 'end' of valid history

    def visit(self, url: str) -> None:
        self.pos += 1
        
        # If we are visiting a new page from the middle of history,
        # we overwrite the next slot.
        if self.pos < len(self.history):
            self.history[self.pos] = url
        else:
            self.history.append(url)
            
        # After a visit, the 'top' is always our current position
        # effectively deleting any forward history.
        self.top = self.pos

    def back(self, steps: int) -> str:
        # Move back, but don't go past index 0
        self.pos = max(0, self.pos - steps)
        return self.history[self.pos]

    def forward(self, steps: int) -> str:
        # Move forward, but don't go past the 'top'
        self.pos = min(self.top, self.pos + steps)
        return self.history[self.pos]


# Your BrowserHistory object will be instantiated and called as such:
# obj = BrowserHistory(homepage)
# obj.visit(url)
# param_2 = obj.back(steps)
# param_3 = obj.forward(steps)

48:T18cd,

class Solution:
    def targetIndices(self, nums: list[int], target: int) -> list[int]:
        count_smaller = 0
        count_target = 0
        
        for num in nums:
            if num < target:
                count_smaller += 1
            elif num == target:
                count_target += 1
                
        # The first index of target will be count_smaller.
        # We generate indices starting from count_smaller for count_target times.
        return [i for i in range(count_smaller, count_smaller + count_target)]

49:T42f,

This problem is very similar to Subarray Sum Equals K, but instead of looking for a specific sum, we are looking for remainders. If the prefix sum up to index i and index j have the same remainder when divided by K, then the subarray between them must be divisible by K.

The Logic:

Prefix Sum with Modulo: As we iterate through the array, maintain a running prefix_sum. Calculate the remainder: remainder = prefix_sum % K.
Handle Negative Remainders: In many programming languages, -1 % 5 results in -1. To ensure we map this correctly, use (prefix_sum % K + K) % K to always get a positive remainder.
Hash Map (Frequency Table): Use a map (or an array of size K) to store how many times each remainder has occurred.
Counting: If a remainder has appeared times before, it means there are n new subarrays ending at the current index that are divisible by K.

4a:T417,class Solution: from collections import defaultdict def subarraysDivByK(self, nums: list[int], k: int) -> int: # count stores the frequency of remainders encountered so far # Initialize with {0: 1} to handle subarrays that are divisible # from the very beginning (index 0). remainder_freq = {0: 1} current_sum = 0 count = 0 for num in nums: current_sum += num # Calculate remainder. Python's % operator handles negative # numbers correctly for this specific logic (e.g., -1 % 5 = 4). remainder = current_sum % k # If we have seen this remainder before, it means there are # subarrays ending here that are divisible by k. if remainder in remainder_freq: count += remainder_freq[remainder] remainder_freq[remainder] += 1 else: remainder_freq[remainder] = 1 return count4b:T23d1,

class Solution:
    from collections import defaultdict
    def subarraysDivByK(self, nums: list[int], k: int) -> int:
        # count stores the frequency of remainders encountered so far
        # Initialize with {0: 1} to handle subarrays that are divisible 
        # from the very beginning (index 0).
        remainder_freq = {0: 1}
        
        current_sum = 0
        count = 0
        
        for num in nums:
            current_sum += num
            
            # Calculate remainder. Python's % operator handles negative 
            # numbers correctly for this specific logic (e.g., -1 % 5 = 4).
            remainder = current_sum % k
            
            # If we have seen this remainder before, it means there are 
            # subarrays ending here that are divisible by k.
            if remainder in remainder_freq:
                count += remainder_freq[remainder]
                remainder_freq[remainder] += 1
            else:
                remainder_freq[remainder] = 1
                
        return count

4c:T15e5,

class Solution:
    def check(self, nums):
        n = len(nums)
        if n <= 2:
            return True

        drops = 0
        for i in range(n):
            if nums[i] > nums[(i + 1) % n]: # Wrap-around comparison
                drops += 1
                if drops > 1:
                    return False
        return True

4d:T14ae,

# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def deleteDuplicates(self, head):
        current = head

        while current and current.next:
            if current.val == current.next.val:
                # Skip the duplicate node
                current.next = current.next.next
            else:
                # Move forward only when values differ
                current = current.next

        return head

4e:T1adf,

class Solution:
    def maxFrequency(self, nums, k):
        nums.sort()

        left = 0
        window_sum = 0
        best = 1

        for right, target in enumerate(nums):
            window_sum += target

            # cost to raise nums[left..right] up to nums[right] (target)
            # = target * window_size - window_sum
            while target * (right - left + 1) - window_sum > k:
                window_sum -= nums[left]
                left += 1

            best = max(best, right - left + 1)

        return best

4f:T1497,

class Solution:
    def firstBadVersion(self, n: int) -> int:
        left, right = 1, n  # invariant: answer is in [left, right]
        while left < right:
            mid = left + (right - left) // 2
            if isBadVersion(mid):
                # mid is bad, so the first bad is at mid or to the left
                right = mid
            else:
                # mid is good, so the first bad must be to the right
                left = mid + 1
        return left

50:T215b,

class Solution:
    def maxSumDivThree(self, nums):
        # dp[i] = max sum such that sum % 3 == i
        # Start with 0 at index 0, and float('-inf') for others
        dp = [0, float('-inf'), float('-inf')]
        
        for num in nums:
            # Use a temporary list to store updates for this current number
            # so we don't use the same number twice in one transition
            next_dp = dp[:]
            
            for current_sum in dp:
                if current_sum == float('-inf'):
                    continue
                    
                new_sum = current_sum + num
                remainder = new_sum % 3
                
                # Update the state if the new sum is larger than what we have
                if new_sum > next_dp[remainder]:
                    next_dp[remainder] = new_sum
            
            dp = next_dp
            
        return dp[0]

51:T3746,

def maxSumDivThree_Greedy(nums):
    total_sum = sum(nums)
    remainder = total_sum % 3
    
    if remainder == 0:
        return total_sum
    
    # Store the two smallest numbers for remainder 1 and remainder 2
    # Initialized to infinity
    rem1 = sorted([n for n in nums if n % 3 == 1])[:2]
    rem2 = sorted([n for n in nums if n % 3 == 2])[:2]
    
    ans = 0
    if remainder == 1:
        # Option A: Remove the smallest rem1
        # Option B: Remove the two smallest rem2
        res1 = total_sum - rem1[0] if len(rem1) >= 1 else 0
        res2 = total_sum - sum(rem2) if len(rem2) >= 2 else 0
        ans = max(res1, res2)
        
    elif remainder == 2:
        # Option A: Remove the smallest rem2
        # Option B: Remove the two smallest rem1
        res1 = total_sum - rem2[0] if len(rem2) >= 1 else 0
        res2 = total_sum - sum(rem1) if len(rem1) >= 2 else 0
        ans = max(res1, res2)
        
    return ans

52:T4af,When we introduce the notion of forming pairs, we're effectively working over the Cartesian product, which we can think of in terms of a matrix. So we have to consider the diagonal. So we throw it all into a heap rather than comparing the immediate pointer-by-pointer thing. We encounter this inflection point where pointers are no longer good enough when the situation involves an additional degree of freedom. We're exploring a frontier, not a path. It's still greedy, but it's a globally greedy strategy rather than a locally greedy one. Two-pointers work when the set of possible next answers can be represented by a constant-size boundary (usually 2 items). So we need a data structure that answers: > "Among all frontier candidates, which sum is smallest right now?" Once our problem's natural candidate is no longer constant-size, that's the inflection point where we typically need something like: - a heap / balanced tree (min/max across many candidates), or - a different global technique (binary search on answer + counting, etc) https://gemini.google.com/app/079f36b2896c994e https://chatgpt.com/g/g-68b11a82d24c81918000aa3a81424e5d-coach/c/695b4169-a420-8328-bb99-bcb44b943a0c53:T5cf,

When we introduce the notion of forming pairs, we're effectively working over the Cartesian product, which we can think of in terms of a matrix.

So we have to consider the diagonal. So we throw it all into a heap rather than comparing the immediate pointer-by-pointer thing.

We encounter this inflection point where pointers are no longer good enough when the situation involves an additional degree of freedom. We're exploring a frontier, not a path.

It's still greedy, but it's a globally greedy strategy rather than a locally greedy one.

Two-pointers work when the set of possible next answers can be represented by a constant-size boundary (usually 2 items).

So we need a data structure that answers:

"Among all frontier candidates, which sum is smallest right now?"

Once our problem's natural candidate is no longer constant-size, that's the inflection point where we typically need something like:

a heap / balanced tree (min/max across many candidates), or
a different global technique (binary search on answer + counting, etc)

https://gemini.google.com/app/079f36b2896c994e

https://chatgpt.com/g/g-68b11a82d24c81918000aa3a81424e5d-coach/c/695b4169-a420-8328-bb99-bcb44b943a0c

54:T3319,

class Solution:
    import heapq
    def kSmallestPairs(self, nums1, nums2, k):
        if not nums1 or not nums2 or k <= 0:
            return []
        
        min_heap = []
        res = []
        
        # 1. Initialize the heap with the first element of nums1 
        # paired with the first few elements of nums2.
        # We only need to go up to min(len(nums1), k)
        for i in range(min(len(nums1), k)):
            heapq.heappush(min_heap, (nums1[i] + nums2[0], i, 0))
            
        # 2. Extract the smallest sum and push the next potential candidate
        while min_heap and len(res) < k:
            current_sum, i, j = heapq.heappop(min_heap)
            res.append([nums1[i], nums2[j]])
            
            # If there is a next element in nums2, pair it with the current nums1[i]
            if j + 1 < len(nums2):
                heapq.heappush(min_heap, (nums1[i] + nums2[j+1], i, j+1))
                
        return res

55:T4f7,class Solution: from collections import deque, defaultdict def sequenceReconstruction(self, nums, sequences): # 1. Build the graph and calculate in-degrees adj = defaultdict(list) in_degree = {num: 0 for num in nums} nodes = set() for seq in sequences: for i in range(len(seq) - 1): u, v = seq[i], seq[i+1] adj[u].append(v) in_degree[v] += 1 nodes.add(u) nodes.add(v) # 2. Initialize the queue with nodes having 0 in-degree queue = deque([node for node in nums if in_degree[node] == 0]) reconstructed_seq = [] # 3. Process the graph while queue: # CRITICAL STEP: If there's more than one choice, the sequence isn't unique if len(queue) > 1: return False curr = queue.popleft() reconstructed_seq.append(curr) for neighbor in adj[curr]: in_degree[neighbor] -= 1 if in_degree[neighbor] == 0: queue.append(neighbor) # 4. Check if the reconstructed sequence matches the target nums return reconstructed_seq == nums56:T3e5d,

class Solution:
    from collections import deque, defaultdict
    def sequenceReconstruction(self, nums, sequences):
        # 1. Build the graph and calculate in-degrees
        adj = defaultdict(list)
        in_degree = {num: 0 for num in nums}
        nodes = set()
        
        for seq in sequences:
            for i in range(len(seq) - 1):
                u, v = seq[i], seq[i+1]
                adj[u].append(v)
                in_degree[v] += 1
                nodes.add(u)
                nodes.add(v)

        # 2. Initialize the queue with nodes having 0 in-degree
        queue = deque([node for node in nums if in_degree[node] == 0])
        reconstructed_seq = []

        # 3. Process the graph
        while queue:
            # CRITICAL STEP: If there's more than one choice, the sequence isn't unique
            if len(queue) > 1:
                return False
            
            curr = queue.popleft()
            reconstructed_seq.append(curr)
            
            for neighbor in adj[curr]:
                in_degree[neighbor] -= 1
                if in_degree[neighbor] == 0:
                    queue.append(neighbor)

        # 4. Check if the reconstructed sequence matches the target nums
        return reconstructed_seq == nums

57:T28fa,

class Solution:
    from collections import deque

    def numOfMinutes(self, n, headID, manager, informTime):
        # 1. Build Adjacency List (Manager -> subordinates)
        adj = [[] for _ in range(n)]
        for employee_id, mgr_id in enumerate(manager):
            if mgr_id != -1:
                adj[mgr_id].append(employee_id)
                
        # 2. BFS: (current_employee, cumulative_time_to_reach_them)
        queue = deque([(headID, 0)])
        max_time = 0
        
        while queue:
            curr_emp, curr_time = queue.popleft()
            
            # Update the total time needed
            max_time = max(max_time, curr_time)
            
            # For every subordinate, the time to reach them is 
            # current time + the time this manager takes to inform everyone
            for subordinate in adj[curr_emp]:
                queue.append((subordinate, curr_time + informTime[curr_emp]))
                
        return max_time

58:T496,

This problem exemplifies the difference between "searching for a value" and "counting a range."

The indices are a proxy for the count.

Sorting turns “which pairs are valid?” into “where is the cutoff index?”, and cutoff indices are countable without enumeration.

We're looking "above" the problem.

This is the inclusion-exclusion principle, or the prefix sum property.

Instead of trying to stay "inside" the narrow window of $[\text{lower}, \text{upper}]$, which is hard to manage with two pointers, we solve the "easier" problem: "How many pairs are less than X?"

Transform a hard condition (“sum in a range”) into something with a monotone boundary (“sum ≤ X”).

Then exploit that boundary to count in bulk.

Finally convert range back via subtraction.

https://gemini.google.com/app/fa4236fb21f99f13

https://chatgpt.com/g/g-68b11a82d24c81918000aa3a81424e5d-coach/c/6957fce7-f670-8326-b624-c02d219440f2

59:T1e24,

class Solution:
    def countFairPairs(self, nums, lower, upper):
        nums.sort()
        
        def countLessEqual(target):
            count = 0
            left = 0
            right = len(nums) - 1
            while left < right:
                if nums[left] + nums[right] <= target:
                    # All pairs between left and right work
                    count += (right - left)
                    left += 1
                else:
                    right -= 1
            return count
        
        return countLessEqual(upper) - countLessEqual(lower - 1)

5a:T2189,

class Solution:
    from functools import cmp_to_key
    def largestNumber(self, nums):
        # Convert numbers to strings
        num_strs = [str(num) for num in nums]
        
        # Custom comparator
        def compare(a, b):
            if a + b > b + a:
                return -1  # a should come first
            else:
                return 1   # b should come first
                
        # Sort with the custom rule
        num_strs.sort(key=cmp_to_key(compare))
        
        # Join strings together
        result = "".join(num_strs)
        
        # Handle the "000" edge case by converting to int and back to str
        # Or simply: return '0' if result[0] == '0' else result
        # if parts and parts[0] == "0":
        #     return "0"
        return str(int(result))

5b:T52b,class Solution: def exclusiveTime(self, n, logs): # ans stores the exclusive time for each function ID ans = [0] * n stack = [] # Stores function IDs # prev_time keeps track of the last timestamp we processed prev_time = 0 for log in logs: # Parse the log string fn_id, type, timestamp = log.split(':') fn_id, timestamp = int(fn_id), int(timestamp) if type == 'start': # If there's a function already running, # add the time elapsed to its exclusive count if stack: ans[stack[-1]] += timestamp - prev_time # Push current function and update current time stack.append(fn_id) prev_time = timestamp else: # type == 'end' # Pop the function and add the duration (inclusive) # +1 because "end at 2" means the function ran through the end of second 2 popped_id = stack.pop() ans[popped_id] += timestamp - prev_time + 1 # Update prev_time to the start of the next second prev_time = timestamp + 1 return ans5c:T2e7d,

class Solution:
    def exclusiveTime(self, n, logs):
        # ans stores the exclusive time for each function ID
        ans = [0] * n
        stack = [] # Stores function IDs
        # prev_time keeps track of the last timestamp we processed
        prev_time = 0
        
        for log in logs:
            # Parse the log string
            fn_id, type, timestamp = log.split(':')
            fn_id, timestamp = int(fn_id), int(timestamp)
            
            if type == 'start':
                # If there's a function already running, 
                # add the time elapsed to its exclusive count
                if stack:
                    ans[stack[-1]] += timestamp - prev_time
                
                # Push current function and update current time
                stack.append(fn_id)
                prev_time = timestamp
                
            else: # type == 'end'
                # Pop the function and add the duration (inclusive)
                # +1 because "end at 2" means the function ran through the end of second 2
                popped_id = stack.pop()
                ans[popped_id] += timestamp - prev_time + 1
                
                # Update prev_time to the start of the next second
                prev_time = timestamp + 1
                
        return ans

5d:T811,For this problem, since any key we pick up stays with us forever and there's no "cost" or "distance" to minimize, this isn't about bitmasking or finding a specific path—it's a straightforward Graph Traversal problem. This is a classic Connected Components problem in a directed graph. The Stack: Represents our "frontier"—the keys we have in our hand but haven't used yet. The Set: Represents our "knowledge"—the rooms we've already unlocked. While this problem is simple, there is one detail to watch out for: Is the graph disconnected? Because we start at Room 0, if the graph is split into two "islands" and the second island contains the keys to itself, we will never reach it. The len(visited) == n check is the most elegant way to catch this. There are two main reasons why the input is usually an array of lists rather than a list of sets, and why we don't bother converting them: 1. Data Redundancy vs. Lookup Speed In a graph traversal (BFS/DFS), we only visit each room once. When we enter a room, you iterate through every key inside it exactly one time to see where we can go next. Iteration: Iterating through a List is slightly faster than iterating through a Set because lists have better memory locality (elements are packed together in an array). Duplicate Keys: Even if a room contains duplicate keys (e.g., rooms[0] = [1, 1, 1]), our visited set handles the logic for us. The if key not in visited check costs $O(1)$. Converting the room's keys into a set just to remove duplicates before we iterate would actually take more time than just letting the visited check skip the duplicates during the crawl. 2. Space Complexity Converting the input to sets requires $O(K)$ extra space (where $K$ is the total number of keys in all rooms combined). For a large input, this essentially doubles the memory footprint of the problem before we've even started solving it. Since we can achieve the result by just reading the lists as they are, the extra space is generally considered "waste." https://gemini.google.com/app/69d26671c5e682eb5e:T8b2,

For this problem, since any key we pick up stays with us forever and there's no "cost" or "distance" to minimize, this isn't about bitmasking or finding a specific path—it's a straightforward Graph Traversal problem.

This is a classic Connected Components problem in a directed graph.

The Stack: Represents our "frontier"—the keys we have in our hand but haven't used yet.

The Set: Represents our "knowledge"—the rooms we've already unlocked.

While this problem is simple, there is one detail to watch out for: Is the graph disconnected? Because we start at Room 0, if the graph is split into two "islands" and the second island contains the keys to itself, we will never reach it. The len(visited) == n check is the most elegant way to catch this.

There are two main reasons why the input is usually an array of lists rather than a list of sets, and why we don't bother converting them:

Data Redundancy vs. Lookup Speed

In a graph traversal (BFS/DFS), we only visit each room once. When we enter a room, you iterate through every key inside it exactly one time to see where we can go next.

Iteration: Iterating through a List is slightly faster than iterating through a Set because lists have better memory locality (elements are packed together in an array).

Duplicate Keys: Even if a room contains duplicate keys (e.g., rooms[0] = [1, 1, 1]), our visited set handles the logic for us. The if key not in visited check costs $O(1)$. Converting the room's keys into a set just to remove duplicates before we iterate would actually take more time than just letting the visited check skip the duplicates during the crawl.

Space Complexity

Converting the input to sets requires $O(K)$ extra space (where $K$ is the total number of keys in all rooms combined). For a large input, this essentially doubles the memory footprint of the problem before we've even started solving it. Since we can achieve the result by just reading the lists as they are, the extra space is generally considered "waste."

https://gemini.google.com/app/69d26671c5e682eb

5f:T1617,

class Solution:
    def canVisitAllRooms(self, rooms):
        visited = {0}
        stack = [0]
        
        while stack:
            current_room = stack.pop()
            for key in rooms[current_room]:
                if key not in visited:
                    visited.add(key)
                    stack.append(key)
                    
        return len(visited) == len(rooms)

60:T1cea,

class Solution:
    def findMaxLength(self, nums):
        # Map to store (prefix_sum: first_index_it_occurred)
        # Initialize with 0: -1 to handle subarrays starting from index 0
        indices = {0: -1}
        max_len = 0
        count = 0
        
        for i, num in enumerate(nums):
            # Treat 0 as -1 and 1 as 1
            count += 1 if num == 1 else -1
            
            if count in indices:
                # Calculate length: current index - first occurrence of this sum
                max_len = max(max_len, i - indices[count])
            else:
                # Only store the first occurrence to maximize the distance/length
                indices[count] = i
                
        return max_len

61:T2841,

class Solution:
    def divide(self, dividend: int, divisor: int) -> int:
        # Constants for 32-bit range
        MAX_INT = 2147483647
        MIN_INT = -2147483648

        # Edge case: Overflow
        if dividend == MIN_INT and divisor == -1:
            return MAX_INT

        # Determine sign
        negative = (dividend < 0) ^ (divisor < 0)
        
        # Use absolute values
        dividend, divisor = abs(dividend), abs(divisor)
        quotient = 0

        # Exponential subtraction
        while dividend >= divisor:
            temp_divisor = divisor
            multiple = 1
            # Double the divisor as much as possible
            while dividend >= (temp_divisor << 1):
                temp_divisor <<= 1
                multiple <<= 1
            
            # Subtract the largest found chunk
            dividend -= temp_divisor
            quotient += multiple

        return -quotient if negative else quotient

62:T432,The pedagogical value of this question is about reducing the problem to a certain state and recognizing that we can perform transformations on a smaller subset of the true data, but it's still reflective of truth because we are able to get to the part of the problem that actually matters. This problem relies on modular arithmetic and the Pigeonhole Principle. We are looking for the smallest n such that a number consisting only of 1s (e.g., 1, 11, 111, ...) is divisible by K. Key Insights: Divisibility Rule: If K is a multiple of 2 or 5, it is impossible for any number ending in 1 to be divisible by K. In these cases, we can immediately return -1. Remainder Cycle: Instead of calculating massive numbers, we can calculate the remainder at each step. Stopping Condition: If we see the same remainder twice, we've entered a cycle and will never reach 0. By the Pigeonhole Principle, since there are only K possible remainders (0 to K-1), if you haven't found a remainder of 0 within K iterations, it doesn't exist. https://gemini.google.com/app/78647b18b5a9237663:T4a1,

The pedagogical value of this question is about reducing the problem to a certain state and recognizing that we can perform transformations on a smaller subset of the true data, but it's still reflective of truth because we are able to get to the part of the problem that actually matters.

This problem relies on modular arithmetic and the Pigeonhole Principle.

We are looking for the smallest n such that a number consisting only of 1s (e.g., 1, 11, 111, ...) is divisible by K.

Key Insights:

Divisibility Rule: If K is a multiple of 2 or 5, it is impossible for any number ending in 1 to be divisible by K. In these cases, we can immediately return -1.

Remainder Cycle: Instead of calculating massive numbers, we can calculate the remainder at each step.

Stopping Condition: If we see the same remainder twice, we've entered a cycle and will never reach 0. By the Pigeonhole Principle, since there are only K possible remainders (0 to K-1), if you haven't found a remainder of 0 within K iterations, it doesn't exist.

https://gemini.google.com/app/78647b18b5a92376

64:T19dc,

class Solution:
    def smallestRepunitDivByK(self, k: int) -> int:
        # If k is even or divisible by 5, no number ending in 1 can be divisible by it
        if k % 2 == 0 or k % 5 == 0:
            return -1
        
        remainder = 0
        # The length of n can at most be k
        for length in range(1, k + 1):
            # Update the remainder: (previous_remainder * 10 + 1) % k
            remainder = (remainder * 10 + 1) % k
            
            # If remainder is 0, we found the smallest n
            if remainder == 0:
                return length
                
        return -1

65:T3647,

class Solution:
    def searchRange(self, nums, target):
        def find_first(nums, target):
            low, high = 0, len(nums) - 1
            first_idx = -1
            while low <= high:
                mid = (low + high) // 2
                if nums[mid] >= target:
                    if nums[mid] == target:
                        first_idx = mid
                    high = mid - 1
                else:
                    low = mid + 1
            return first_idx

        def find_last(nums, target):
            low, high = 0, len(nums) - 1
            last_idx = -1
            while low <= high:
                mid = (low + high) // 2
                if nums[mid] <= target:
                    if nums[mid] == target:
                        last_idx = mid
                    low = mid + 1
                else:
                    high = mid - 1
            return last_idx

        return [find_first(nums, target), find_last(nums, target)]

66:T15f5,

class Solution:
    def searchRange(self, nums, target):
        left = bisect.bisect_left(nums, target)
        right = bisect.bisect_right(nums, target) - 1
        
        # Check if left is within bounds and actually points to the target
        if left < len(nums) and nums[left] == target:
            return [left, right]
        
        return [-1, -1]

67:T429,

I'm just putting this here for posterity. We should expect to not see this, but there're a lot of thoughts I've been having about the essence of programming, and what it means to seek truth.

I've found that two axes can relatively nicely distil what I've been thinking:

We want to reduce ambiguity — this means exhaustively mapping out the cases. Following the implicit structure and making it explicit, or imposing a known algorithm or structure to make a problem conform to a known solution.
We want to reduce complexity — this comes in the form of identifying places where we can make use of simplicity. What's true is often more simple, relatively speaking.

https://gemini.google.com/app/996e78a63b751252

https://chatgpt.com/g/g-68b11a82d24c81918000aa3a81424e5d-coach/c/694c8197-65c8-8328-92b7-5f4cc6378ba5

68:T58e,class Solution: def countRangeSum(self, nums: list[int], lower: int, upper: int) -> int: # Step 1: Generate Prefix Sums # prefix_sums[i] is the sum of nums[0...i-1] # We include 0 to handle ranges starting from the very first element S = [0] for x in nums: S.append(S[-1] + x) def count_while_sorting(left, right): if right - left <= 1: return 0 mid = (left + right) // 2 # Recursively count pairs in the left half and right half count = count_while_sorting(left, mid) + count_while_sorting(mid, right) # Step 2: The "Magic" Count Step # For every i in the left half, find the range [j_low, j_high) # in the right half such that lower <= S[j] - S[i] <= upper j_low = j_high = mid for i in range(left, mid): while j_low < right and S[j_low] - S[i] < lower: j_low += 1 while j_high < right and S[j_high] - S[i] <= upper: j_high += 1 count += (j_high - j_low) # Step 3: Standard Merge Step # This keeps the prefix sum array sorted for the next level of recursion S[left:right] = sorted(S[left:right]) return count return count_while_sorting(0, len(S))69:T431f,

class Solution:
    def countRangeSum(self, nums: list[int], lower: int, upper: int) -> int:
        # Step 1: Generate Prefix Sums
        # prefix_sums[i] is the sum of nums[0...i-1]
        # We include 0 to handle ranges starting from the very first element
        S = [0]
        for x in nums:
            S.append(S[-1] + x)

        def count_while_sorting(left, right):
            if right - left <= 1:
                return 0
            
            mid = (left + right) // 2
            # Recursively count pairs in the left half and right half
            count = count_while_sorting(left, mid) + count_while_sorting(mid, right)
            
            # Step 2: The "Magic" Count Step
            # For every i in the left half, find the range [j_low, j_high) 
            # in the right half such that lower <= S[j] - S[i] <= upper
            j_low = j_high = mid
            for i in range(left, mid):
                while j_low < right and S[j_low] - S[i] < lower:
                    j_low += 1
                while j_high < right and S[j_high] - S[i] <= upper:
                    j_high += 1
                count += (j_high - j_low)
            
            # Step 3: Standard Merge Step
            # This keeps the prefix sum array sorted for the next level of recursion
            S[left:right] = sorted(S[left:right])
            return count

        return count_while_sorting(0, len(S))

6a:T3094,

class Solution:
    def reversePairs(self, nums: list[int]) -> int:
        def merge_sort(start, end):
            if start >= end:
                return 0
            
            mid = (start + end) // 2
            count = merge_sort(start, mid) + merge_sort(mid + 1, end)
            
            # --- Counting Step ---
            # Use two pointers to count reverse pairs between the two sorted halves
            j = mid + 1
            for i in range(start, mid + 1):
                while j <= end and nums[i] > 2 * nums[j]:
                    j += 1
                count += (j - (mid + 1))
            
            # --- Standard Merge Step ---
            nums[start:end+1] = sorted(nums[start:end+1])
            return count

        return merge_sort(0, len(nums) - 1)

6b:T66f,class Solution: def maxScoreWords(self, words, letters, score): # 1. Count our available resources letter_counts = Counter(letters) def get_word_score(word): """Helper to calculate score of a word; returns 0 if impossible.""" s = 0 for char in word: s += score[ord(char) - ord('a')] return s def can_form(word, current_counts): """Check if we have enough letters left to form the word.""" word_counts = Counter(word) for char, count in word_counts.items(): if current_counts[char] < count: return False return True def backtrack(index, current_counts): if index == len(words): return 0 # Option 1: Skip the current word max_s = backtrack(index + 1, current_counts) # Option 2: Try to take the current word word = words[index] if can_form(word, current_counts): # "Pay" the letters for char in word: current_counts[char] -= 1 # Add word score + recurse take_score = get_word_score(word) + backtrack(index + 1, current_counts) max_s = max(max_s, take_score) # "Backtrack": Put the letters back for other branches for char in word: current_counts[char] += 1 return max_s return backtrack(0, letter_counts)6c:T4443,

class Solution:
    def maxScoreWords(self, words, letters, score):
        # 1. Count our available resources
        letter_counts = Counter(letters)
        
        def get_word_score(word):
            """Helper to calculate score of a word; returns 0 if impossible."""
            s = 0
            for char in word:
                s += score[ord(char) - ord('a')]
            return s

        def can_form(word, current_counts):
            """Check if we have enough letters left to form the word."""
            word_counts = Counter(word)
            for char, count in word_counts.items():
                if current_counts[char] < count:
                    return False
            return True

        def backtrack(index, current_counts):
            if index == len(words):
                return 0
            
            # Option 1: Skip the current word
            max_s = backtrack(index + 1, current_counts)
            
            # Option 2: Try to take the current word
            word = words[index]
            if can_form(word, current_counts):
                # "Pay" the letters
                for char in word:
                    current_counts[char] -= 1
                
                # Add word score + recurse
                take_score = get_word_score(word) + backtrack(index + 1, current_counts)
                max_s = max(max_s, take_score)
                
                # "Backtrack": Put the letters back for other branches
                for char in word:
                    current_counts[char] += 1
                    
            return max_s

        return backtrack(0, letter_counts)

6d:T27d2,

class Solution:
    def intToRoman(self, num: int) -> str:
        # Define the mapping in descending order
        roman_map = [
            (1000, "M"), (900, "CM"), (500, "D"), (400, "CD"),
            (100, "C"), (90, "XC"), (50, "L"), (40, "XL"),
            (10, "X"), (9, "IX"), (5, "V"), (4, "IV"), (1, "I")
        ]
        
        result = []
        
        for value, symbol in roman_map:
            # Check how many times the current symbol fits into num
            count = num // value
            if count > 0:
                result.append(symbol * count)
                # Reduce num by the amount we just converted
                num %= value
                
        return "".join(result)

6e:T1fd1,

class Solution:
    def binaryTreePaths(self, root):
        def dfs(node, path_list):
            if not node:
                return
            
            # 1. Add current value to the list
            path_list.append(str(node.val))
            
            # 2. Check for leaf
            if not node.left and not node.right:
                res.append("->".join(path_list))
            else:
                # 3. Recurse down
                dfs(node.left, path_list)
                dfs(node.right, path_list)
            
            # 4. BACKTRACK: Remove the node so it doesn't leak into the next path
            path_list.pop()

        res = []
        dfs(root, [])
        return res

6f:T451,

We have to take a leap of faith with induction. The dfs function promises that when it finishes executing, the prefix_sums map will be in the exact same state it was in before the function started.

We add one to the map, and we propagate it down to the branches. And we pop it off. And we trust that the children follow the instructions of the parent, because they hold the exact instruction sets.

We end up with a net 0 prefix_sums map at the end.

Other than the 0: 1 entry. That's the ground truth.

There's a lot of semantic meaning in the +1 and -1. These aren't merely isolated operations. These are the physical boundaries of the subproblem's scope.

Once we trust the induction, the code is trivial. We don't have to worry about "What if the left child changes the map?" because we know that the post-condition prevents that. We can focus entirely on the current node's responsibility:

Check the mile markers above you.
Leave your own mile marker for those above you.
Clean up before you leave.

70:T56e,# Definition for a binary tree node. # class TreeNode: # def __init__(self, val=0, left=None, right=None): # self.val = val # self.left = left # self.right = right class Solution: def pathSum(self, root: Optional[TreeNode], targetSum: int) -> int: # map stores {prefix_sum : count} # We initialize with {0: 1} to account for paths that # equal targetSum starting exactly from the root. prefix_sums = {0: 1} def dfs(node, current_sum): if not node: return 0 # Update the running prefix sum current_sum += node.val # If (current_sum - targetSum) exists in our map, # it means a valid path ends at this node. num_paths = prefix_sums.get(current_sum - targetSum, 0) # Add current_sum to the map for child nodes to use prefix_sums[current_sum] = prefix_sums.get(current_sum, 0) + 1 # Explore children num_paths += dfs(node.left, current_sum) # These don't share state because we backtrack. num_paths += dfs(node.right, current_sum) # Backtracking: remove the current sum from the map # so it won't affect other branches prefix_sums[current_sum] -= 1 return num_paths return dfs(root, 0) 71:T3216,

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def pathSum(self, root: Optional[TreeNode], targetSum: int) -> int:
        # map stores {prefix_sum : count}
        # We initialize with {0: 1} to account for paths that
        # equal targetSum starting exactly from the root.
        prefix_sums = {0: 1}

        def dfs(node, current_sum):
            if not node:
                return 0
            
            # Update the running prefix sum
            current_sum += node.val

            # If (current_sum - targetSum) exists in our map, 
            # it means a valid path ends at this node.
            num_paths = prefix_sums.get(current_sum - targetSum, 0)

            # Add current_sum to the map for child nodes to use
            prefix_sums[current_sum] = prefix_sums.get(current_sum, 0) + 1

            # Explore children
            num_paths += dfs(node.left, current_sum)
            # These don't share state because we backtrack.
            num_paths += dfs(node.right, current_sum)

            # Backtracking: remove the current sum from the map
            # so it won't affect other branches
            prefix_sums[current_sum] -= 1

            return num_paths
        return dfs(root, 0)

72:T19ca,

class Solution:
    def hasPathSum(self, root: Optional[TreeNode], targetSum: int) -> bool:
        if not root:
            return False
        
        # If it's a leaf, it has to be the value
        if not root.left and not root.right:
            return root.val == targetSum
        
        remaining = targetSum - root.val
        return self.hasPathSum(root.left, remaining) or self.hasPathSum(root.right, remaining)

73:T429,The most elegant way to solve this is using Post-Order Traversal (Bottom-Up). Set Optimization: Convert to_delete to a set for O(1) lookup time. Recursion Logic: We define a helper function that returns None if the current node should be deleted (telling the parent to sever the link) or returns the node itself if it should stay. Identifying New Roots: If a node is marked for deletion, its children (if they exist and haven't been deleted) automatically become the roots of new trees. We add them to our result list immediately before deleting the current node. Why write it like this? Single Responsibility: Each call to process(node) has one job: "Clean my subtrees and tell my parent if I still exist." Implicit Logic: You don't need to pass state (is_root) down. The state is handled naturally by the call stack. The "Snip" is Automatic: By setting node.left = process(node.left), the parent automatically loses the reference to a deleted child without any if/else logic needed inside the parent's block. https://gemini.google.com/app/2d978edfffefdc4574:T49e,

The most elegant way to solve this is using Post-Order Traversal (Bottom-Up).

Set Optimization: Convert to_delete to a set for O(1) lookup time.

Recursion Logic: We define a helper function that returns None if the current node should be deleted (telling the parent to sever the link) or returns the node itself if it should stay.

Identifying New Roots: If a node is marked for deletion, its children (if they exist and haven't been deleted) automatically become the roots of new trees. We add them to our result list immediately before deleting the current node.

Why write it like this?

Single Responsibility: Each call to process(node) has one job: "Clean my subtrees and tell my parent if I still exist."

Implicit Logic: You don't need to pass state (is_root) down. The state is handled naturally by the call stack.

The "Snip" is Automatic: By setting node.left = process(node.left), the parent automatically loses the reference to a deleted child without any if/else logic needed inside the parent's block.

https://gemini.google.com/app/2d978edfffefdc45

75:T44c,class Solution: def delNodes(self, root: Optional[TreeNode], to_delete: List[int]) -> List[TreeNode]: # 1. Use a set for O(1) lookups delete_set = set(to_delete) forest = [] def process(node): if not node: return None # 2. Process children first (Bottom-Up) # This ensures children are already handled if they needed deleting node.left = process(node.left) node.right = process(node.right) # 3. Handle the current node if node.val in delete_set: # If this node is gone, its surviving children are now independent roots if node.left: forest.append(node.left) if node.right: forest.append(node.right) return None # "Delete" this node by returning None to its parent return node # Keep this node # 4. Kick off the process and check if the original root survived if process(root): forest.append(root) return forest76:T2d79,

class Solution:
    def delNodes(self, root: Optional[TreeNode], to_delete: List[int]) -> List[TreeNode]:
        # 1. Use a set for O(1) lookups
        delete_set = set(to_delete)
        forest = []

        def process(node):
            if not node:
                return None
            
            # 2. Process children first (Bottom-Up)
            # This ensures children are already handled if they needed deleting
            node.left = process(node.left)
            node.right = process(node.right)

            # 3. Handle the current node
            if node.val in delete_set:
                # If this node is gone, its surviving children are now independent roots
                if node.left: forest.append(node.left)
                if node.right: forest.append(node.right)
                return None # "Delete" this node by returning None to its parent
            
            return node # Keep this node

        # 4. Kick off the process and check if the original root survived
        if process(root):
            forest.append(root)
            
        return forest

77:T1ade,

class Solution:
    def peakIndexInMountainArray(self, arr: List[int]) -> int:
        left, right = 0, len(arr) - 1
        while left < right:
            mid = (left + right) // 2

            # Check if we're on an uphill or a downhill
            if arr[mid] < arr[mid + 1]:
                # We're on the left side of the peak! Move right
                left = mid + 1
            else:
                # We're either at the peak or on the right side! Move left
                # We can't do mid - 1 because mid might be the peak itself
                right = mid
        return left

78:T4de,It's time for Dynamic Programming! BFS is great for finding the shortest path in an unweighted graph, or Dijkstra for weighted graphs. However, since the movement here is strictly row-by-row, that makes it a Directed Acyclic Graph, which means that using a full graph traversal algorithm is overkill. We simply need to accumulate results. Greedy won't work because a small number now might lead us to a massive number later. Instead, we can use tabulation. We can modify the matrix in place (or create a copy). For every cell, we can ask "If I had to end up at this cell, what is the minimum cost to get here?" The Recurrence Relation: To arrive at matrix[row][col], we must have come from one of three places in the row-1: 1. Directly above (row-1, col) 2. Diagonal left: (row-1, col-1) 3. Directly right: (row-1, col+1) So the value for the current cell becomes: matrix[r][c] = matrix[r][c] + min(above, left_diag, right_diag). This will be O(N^2), since we're visiting every cell once. The space is O(1) if we modify the input matrix, and O(N^2) if we make a copy. At every cell, we're saying: "I don't care about the history of how I got here; I only care about the best score I could possibly have to arrive at this specific spot.79:T53e,

It's time for Dynamic Programming!

BFS is great for finding the shortest path in an unweighted graph, or Dijkstra for weighted graphs. However, since the movement here is strictly row-by-row, that makes it a Directed Acyclic Graph, which means that using a full graph traversal algorithm is overkill.

We simply need to accumulate results.

Greedy won't work because a small number now might lead us to a massive number later.

Instead, we can use tabulation. We can modify the matrix in place (or create a copy). For every cell, we can ask "If I had to end up at this cell, what is the minimum cost to get here?"

The Recurrence Relation:

To arrive at matrix[row][col], we must have come from one of three places in the row-1:

Directly above (row-1, col)
Diagonal left: (row-1, col-1)
Directly right: (row-1, col+1)

So the value for the current cell becomes:

matrix[r][c] = matrix[r][c] + min(above, left_diag, right_diag).

This will be O(N^2), since we're visiting every cell once. The space is O(1) if we modify the input matrix, and O(N^2) if we make a copy.

At every cell, we're saying: "I don't care about the history of how I got here; I only care about the best score I could possibly have to arrive at this specific spot.

7a:T2e71,

class Solution:
    def minFallingPathSum(self, matrix: List[List[int]]) -> int:
        rows = len(matrix)
        cols = len(matrix[0])

        # Start from the second row (index 1)
        for r in range(1, rows):
            for c in range(cols):
                # 1) Directly above
                best_above = matrix[r-1][c]

                # 2) Diagonal left maybe
                if c > 0:
                    best_above = min(best_above, matrix[r-1][c-1])
                
                # 3) Diagonal right maybe
                if c < cols - 1:
                    best_above = min(best_above, matrix[r-1][c+1])
                
                # Update current cell
                matrix[r][c] += best_above
        return min(matrix[-1])

7b:T1eed,

class Solution:
    def findMin(self, nums: List[int]) -> int:
        left, right = 0, len(nums) - 1
        while left < right:
            mid = (left + right) // 2

            if nums[mid] < nums[right]:
                # Minimum is in [left..mid]
                right = mid
            elif nums[mid] > nums[right]:
                # Minimum is in (mid..right]
                left = mid + 1
            else:
                # Equal, so shrink the window
                right -= 1
        return nums[left]

7c:T8ea,class Solution: def longestDupSubstring(self, s: str) -> str: # --- PART 1: The Setup (Math constants) --- nums = [ord(c) - ord('a') for c in s] MOD = 2**63 - 1 # Big prime number to avoid collisions BASE = 26 # 26 lowercase letters n = len(s) # --- PART 2: The "Check" Function (Rabin-Karp) --- # Returns start index if found, -1 if not def search(length): # 1. Compute hash of the FIRST window (0 to length-1) current_hash = 0 for i in range(length): current_hash = (current_hash * BASE + nums[i]) % MOD seen = {current_hash} # Constant needed to remove the leading character: BASE^(L-1) # This is the value of the "highest place" in our number system highest_pow = pow(BASE, length - 1, MOD) # 2. Rolling Update (The "O(1)" magic) for i in range(1, n - length + 1): # Logic: (Current - Leading_Char_Value) * Base + New_Char_Value # Remove leading char (previous window's first char) # We add MOD to ensure the result isn't negative before modulo current_hash = (current_hash - nums[i-1] * highest_pow) % MOD # Shift left and add new trailing char current_hash = (current_hash * BASE + nums[i + length - 1]) % MOD if current_hash in seen: return i seen.add(current_hash) return -1 # --- PART 3: The Driver (Binary Search on Answer) --- # Exactly the same logic as Koko Eating Bananas / Ship Packages left, right = 1, n - 1 start_index = -1 while left <= right: mid = (left + right) // 2 found = search(mid) if found != -1: start_index = found # Valid! Save it. left = mid + 1 # Try for longer else: right = mid - 1 # Impossible. Try shorter. return s[start_index : start_index + (left - 1)] if start_index != -1 else ""7d:T5ab6,

class Solution:
    def longestDupSubstring(self, s: str) -> str:
        # --- PART 1: The Setup (Math constants) ---
        nums = [ord(c) - ord('a') for c in s]
        MOD = 2**63 - 1  # Big prime number to avoid collisions
        BASE = 26        # 26 lowercase letters
        n = len(s)

        # --- PART 2: The "Check" Function (Rabin-Karp) ---
        # Returns start index if found, -1 if not
        def search(length):
            # 1. Compute hash of the FIRST window (0 to length-1)
            current_hash = 0
            for i in range(length):
                current_hash = (current_hash * BASE + nums[i]) % MOD
            
            seen = {current_hash}
            
            # Constant needed to remove the leading character: BASE^(L-1)
            # This is the value of the "highest place" in our number system
            highest_pow = pow(BASE, length - 1, MOD)
            
            # 2. Rolling Update (The "O(1)" magic)
            for i in range(1, n - length + 1):
                # Logic: (Current - Leading_Char_Value) * Base + New_Char_Value
                
                # Remove leading char (previous window's first char)
                # We add MOD to ensure the result isn't negative before modulo
                current_hash = (current_hash - nums[i-1] * highest_pow) % MOD
                
                # Shift left and add new trailing char
                current_hash = (current_hash * BASE + nums[i + length - 1]) % MOD
                
                if current_hash in seen:
                    return i
                seen.add(current_hash)
                
            return -1

        # --- PART 3: The Driver (Binary Search on Answer) ---
        # Exactly the same logic as Koko Eating Bananas / Ship Packages
        left, right = 1, n - 1
        start_index = -1
        
        while left <= right:
            mid = (left + right) // 2
            found = search(mid)
            
            if found != -1:
                start_index = found # Valid! Save it.
                left = mid + 1      # Try for longer
            else:
                right = mid - 1     # Impossible. Try shorter.
                
        return s[start_index : start_index + (left - 1)] if start_index != -1 else ""

7e:T4df,Eventually, all monotonic stack problems have a panacea which consists of two questions: "What happens if it's ascending? What happens if it's descending?" But it's not just about the two extremes. Sure, they illuminate the behavioural boundaries, but it's also what we're doing at every step of the process. In fact, it's all we're doing. It's the only two operations that the algorithm performs, over and over. The switch flips. It is a mechanism to enforce a specific slope by eagerly destroying anything that violates it. Instead of looking ahead to find the smallest, we can look at the neighbour, and iterate through the string only once. We use a stack to keep track of the digits we represent "keeping." The state of the world when the loop finishes is that the stack is in ascending (non-decreasing) order. If the loop finishes and we still have k > 0, it means we never found a "peak" to chop off. We can't promote the second number, or anything at all. We have to start chopping the back. That's the fat to trim. There's a really funny way of seeing this problem as a really intense NIMBY Mayor where you're given k Demolition Permits to destroy tall buildings. We want a really low building, and we want to have a view, and sun. 7f:T50e,

Eventually, all monotonic stack problems have a panacea which consists of two questions: "What happens if it's ascending? What happens if it's descending?"

But it's not just about the two extremes. Sure, they illuminate the behavioural boundaries, but it's also what we're doing at every step of the process. In fact, it's all we're doing. It's the only two operations that the algorithm performs, over and over. The switch flips.

It is a mechanism to enforce a specific slope by eagerly destroying anything that violates it.

Instead of looking ahead to find the smallest, we can look at the neighbour, and iterate through the string only once. We use a stack to keep track of the digits we represent "keeping."

The state of the world when the loop finishes is that the stack is in ascending (non-decreasing) order.

If the loop finishes and we still have k > 0, it means we never found a "peak" to chop off. We can't promote the second number, or anything at all. We have to start chopping the back. That's the fat to trim.

There's a really funny way of seeing this problem as a really intense NIMBY Mayor where you're given k Demolition Permits to destroy tall buildings. We want a really low building, and we want to have a view, and sun.

80:T2337,

class Solution:
    def removeKdigits(self, num: str, k: int) -> str:
        stack = []

        for digit in num:
            # While we have k to spend, the stack isn't empty,
            # and the previous digit is bigger than the current one:
            while k > 0 and stack and stack[-1] > digit:
                stack.pop()
                k -= 1
            stack.append(digit)
        # Edge Case 2: If the numbers were already ascending (e.g. '12345', k=2),
        # we might still have k left to remove. Remove from the end.
        if k > 0:
            stack = stack[:-k]
        # Join the stack to form the string
        result = ''.join(stack)

        # Edge case 3: remove leading zeroes
        result = result.lstrip('0')

        # Edge case 4: If the string is empty (we removed everything), return '0'
        return result if result else '0'

81:T4a0,

Since you can make at most two transactions and they cannot overlap, imagine splitting the array of prices into two parts at a specific day i.

The problem can be split into two sub-problems. If we are allowed two transactions, there must be some dividing line, say day i, where:

Transaction 1 happens on or before day i.
Transaction 2 happens on or after day i.

By iterating through every possible split point i and asking "What is the max profit on the left + max profit on the right?", you guarantee finding the global maximum.

To transition into the O(1) space solution, we can think:

Since Transaction 2 occurs after Transaction 1, we don't need to store the entire history of Transaction 1. We can just calculate the current state of T1 and feed it directly into the cost basis of T2 in the same loop. This removes the need for the array.

https://gemini.google.com/app/b5d7100d95f3bfaa

https://aistudio.google.com/prompts/1Lzk91exDfjXevJyCuhachpqQvj3eabzy

82:T6a6,class Solution: def maxProfit(self, prices: List[int]) -> int: n = len(prices) if n < 2: return 0 # 1. Forward Pass (Pre-compute Max Profit for Transaction 1) # left_profits[i] = max profit achievable from day 0 to i left_profits = [0] * n min_price = prices[0] for i in range(1, n): # Capture the lowest buy price seen so far min_price = min(min_price, prices[i]) # Profit is either what we had before, or selling today left_profits[i] = max(left_profits[i-1], prices[i] - min_price) # 2. Backward Pass (Calculate Transaction 2 & Combine) # We don't need an array for right_profits; just a running max. max_price = prices[-1] right_profit = 0 max_total_profit = 0 # Iterate backwards to find the best second transaction for i in range(n - 1, -1, -1): # Capture the highest sell price seen from the right max_price = max(max_price, prices[i]) # Profit for T2 if we bought on day i and sold at max_price current_right_profit = max_price - prices[i] # Keep track of the best T2 profit found from i to the end right_profit = max(right_profit, current_right_profit) # COMBINE: # T1 profit ending by day i (left_profits[i]) # + T2 profit starting from day i (right_profit) current_total = left_profits[i] + right_profit max_total_profit = max(max_total_profit, current_total) return max_total_profit83:T4350,

class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        n = len(prices)
        if n < 2:
            return 0
        
        # 1. Forward Pass (Pre-compute Max Profit for Transaction 1)
        # left_profits[i] = max profit achievable from day 0 to i
        left_profits = [0] * n
        min_price = prices[0]
        
        for i in range(1, n):
            # Capture the lowest buy price seen so far
            min_price = min(min_price, prices[i])
            # Profit is either what we had before, or selling today
            left_profits[i] = max(left_profits[i-1], prices[i] - min_price)
            
        # 2. Backward Pass (Calculate Transaction 2 & Combine)
        # We don't need an array for right_profits; just a running max.
        max_price = prices[-1]
        right_profit = 0
        max_total_profit = 0
        
        # Iterate backwards to find the best second transaction
        for i in range(n - 1, -1, -1):
            # Capture the highest sell price seen from the right
            max_price = max(max_price, prices[i])
            
            # Profit for T2 if we bought on day i and sold at max_price
            current_right_profit = max_price - prices[i]
            
            # Keep track of the best T2 profit found from i to the end
            right_profit = max(right_profit, current_right_profit)
            
            # COMBINE:
            # T1 profit ending by day i (left_profits[i])
            # + T2 profit starting from day i (right_profit)
            current_total = left_profits[i] + right_profit
            max_total_profit = max(max_total_profit, current_total)
            
        return max_total_profit

84:T439,We can optimize the space by realizing this is essentially a reduction operation. We can just maintain the 'accumulated' best state for the two transactions as we iterate, without saving the history. We don't need the whole history of prices; we just need to carry the state of our wallet — specifically our costs and profits — forward one day at a time. We can model the process as a sequence of 4 specific actions (states) that must happen in order: Buy 1: Minimize the cost of the first stock. Sell 1: Maximize the profit of the first transaction (Price - Buy 1). Buy 2: Minimize the effective cost of the second stock. The "effective cost" is the current price minus the profit gained from the first transaction. Sell 2: Maximize the total profit (Price - Effective Buy 2). This order allows a single price spike to ripple through all transactions on the same day (Buy 1, Sell 1, Buy 2, Sell 2 all at once), which is mathematically valid (resulting in 0 extra profit) and ensures we never miss a peak. We iterate through the prices once, updating these four variables.85:T469,

We can optimize the space by realizing this is essentially a reduction operation. We can just maintain the 'accumulated' best state for the two transactions as we iterate, without saving the history. We don't need the whole history of prices; we just need to carry the state of our wallet — specifically our costs and profits — forward one day at a time.

We can model the process as a sequence of 4 specific actions (states) that must happen in order:

Buy 1: Minimize the cost of the first stock.
Sell 1: Maximize the profit of the first transaction (Price - Buy 1).

Buy 2: Minimize the effective cost of the second stock. The "effective cost" is the current price minus the profit gained from the first transaction.
Sell 2: Maximize the total profit (Price - Effective Buy 2).

This order allows a single price spike to ripple through all transactions on the same day (Buy 1, Sell 1, Buy 2, Sell 2 all at once), which is mathematically valid (resulting in 0 extra profit) and ensures we never miss a peak.

We iterate through the prices once, updating these four variables.

86:T2302,

class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        # Initialize costs to infinity and profits to 0
        buy1 = float('inf') # Cost of first stock
        profit1 = 0 # Profit after selling first stock

        buy2 = float('inf') # Cost of second stock (reinvested)
        profit2 = 0 # Profit after selling second stock

        for price in prices:
            # 1) Minimize the cost of the first stock
            buy1 = min(buy1, price)

            # 2) Maximize profit of the first transaction
            profit1 = max(profit1, price - buy1)

            # 3) Minimize cost of second stock
            # (Treat previous profit as a discount on the second buy price)
            buy2 = min(buy2, price - profit1)

            # 4) Maximize total profit after second transaction
            profit2 = max(profit2, price - buy2)
        return profit2

87:T786,We will split the problem into two explicit, easy-to-read phases: Pre-processing: Flatten the confusing "radius" logic into a simple list of "Start to End" intervals. Simulation: Use a while loop to simulate the process of "standing" at our current limit and looking for the best tap to extend our reach. Phase 1: Flattening the Map The input gives us taps with a center and a radius. This is messy to think about. We really just care about: "If I start watering at point X, how far right can I go?" We can build a simple array max_reach where the index is the start point of the water, and the value is the end point. # ranges = [3, 4, 1, 1, 0, 0], n = 5 # Tap at index 0 (range 3) covers [-3, 3] -> clipped to [0, 3] # Tap at index 1 (range 4) covers [-3, 5] -> clipped to [0, 5] # We transform this into: # At start point 0, the best reach is index 5. Phase 2: The Simulation Loop Instead of a confusing for loop with hidden state changes, let's use a while loop that reads like English: State: We currently have water up to current_end. Action: Look at all the intervals that start within our current watered area. Decision: Pick the one that extends the furthest to the right. Update: Set current_end to that new furthest point and increment the tap count. This is better because: There's no magic i == current_end: The state update happens explicitly at the end of the while block (current_end = next_end). You can see exactly when we decide to add a tap. Separate Scanning Logic: The inner while loop makes it obvious that we are "scanning" the available options within our current reach. Gap Detection: The condition if next_end == current_end clearly signifies "we checked all options and couldn't move forward," which means the garden is impossible to water. This solution is still O(N) time complexity (since index_to_check only moves from 0 to n once), but it prioritizes readability over code golf.88:T812,

We will split the problem into two explicit, easy-to-read phases:

Pre-processing: Flatten the confusing "radius" logic into a simple list of "Start to End" intervals.

Simulation: Use a while loop to simulate the process of "standing" at our current limit and looking for the best tap to extend our reach.

Phase 1: Flattening the Map
The input gives us taps with a center and a radius. This is messy to think about. We really just care about: "If I start watering at point X, how far right can I go?"

We can build a simple array max_reach where the index is the start point of the water, and the value is the end point.

ranges = [3, 4, 1, 1, 0, 0], n = 5

Tap at index 0 (range 3) covers [-3, 3] -> clipped to [0, 3]

Tap at index 1 (range 4) covers [-3, 5] -> clipped to [0, 5]

We transform this into:

At start point 0, the best reach is index 5.

Phase 2: The Simulation Loop
Instead of a confusing for loop with hidden state changes, let's use a while loop that reads like English:

State: We currently have water up to current_end.

Action: Look at all the intervals that start within our current watered area.

Decision: Pick the one that extends the furthest to the right.

Update: Set current_end to that new furthest point and increment the tap count.

This is better because:
There's no magic i == current_end: The state update happens explicitly at the end of the while block (current_end = next_end). You can see exactly when we decide to add a tap.

Separate Scanning Logic: The inner while loop makes it obvious that we are "scanning" the available options within our current reach.

Gap Detection: The condition if next_end == current_end clearly signifies "we checked all options and couldn't move forward," which means the garden is impossible to water.

This solution is still O(N) time complexity (since index_to_check only moves from 0 to n once), but it prioritizes readability over code golf.

89:T630,class Solution: def minTaps(self, n: int, ranges: list[int]) -> int: # --- PHASE 1: Pre-process Taps into Intervals --- # max_reach[i] means: "If a tap starts covering at index 'i', # what is the furthest index it reaches?" max_reach = [0] * (n + 1) for i, r in enumerate(ranges): # Convert (center, radius) to (start, end) start = max(0, i - r) end = min(n, i + r) # We only care about the tap that reaches the furthest from this start point max_reach[start] = max(max_reach[start], end) # --- PHASE 2: Simulation (Stitching) --- taps = 0 current_end = 0 next_end = 0 index_to_check = 0 # While we haven't watered the whole garden while current_end < n: # Scan all possible intervals that start within our current watered range # to find the one that pushes 'next_end' the furthest. while index_to_check <= current_end: next_end = max(next_end, max_reach[index_to_check]) index_to_check += 1 # If we couldn't find ANY tap to extend our reach, we are stuck. # This implies a gap in coverage. if next_end == current_end: return -1 # We found the best tap. "Turn it on." # Our watered area now extends to 'next_end'. current_end = next_end taps += 1 return taps8a:T37da,

class Solution:
    def minTaps(self, n: int, ranges: list[int]) -> int:
        # --- PHASE 1: Pre-process Taps into Intervals ---
        # max_reach[i] means: "If a tap starts covering at index 'i', 
        # what is the furthest index it reaches?"
        max_reach = [0] * (n + 1)
        
        for i, r in enumerate(ranges):
            # Convert (center, radius) to (start, end)
            start = max(0, i - r)
            end = min(n, i + r)
            
            # We only care about the tap that reaches the furthest from this start point
            max_reach[start] = max(max_reach[start], end)

        # --- PHASE 2: Simulation (Stitching) ---
        taps = 0
        current_end = 0
        next_end = 0
        index_to_check = 0
        
        # While we haven't watered the whole garden
        while current_end < n:
            
            # Scan all possible intervals that start within our current watered range
            # to find the one that pushes 'next_end' the furthest.
            while index_to_check <= current_end:
                next_end = max(next_end, max_reach[index_to_check])
                index_to_check += 1
                
            # If we couldn't find ANY tap to extend our reach, we are stuck.
            # This implies a gap in coverage.
            if next_end == current_end:
                return -1
            
            # We found the best tap. "Turn it on."
            # Our watered area now extends to 'next_end'.
            current_end = next_end
            taps += 1
            
        return taps

8b:T637,from typing import List class Solution: def minTaps(self, n: int, ranges: List[int]) -> int: # Array to store the furthest point reachable from each index. # max_reach[i] = max end point for an interval starting at i max_reach = [0] * (n + 1) # 1. Preprocessing: Convert tap ranges into "jumps" for i, r in enumerate(ranges): # Calculate the effective bounds of the tap start = max(0, i - r) end = min(n, i + r) # We want to know: "If I start at `start`, how far can I go?" max_reach[start] = max(max_reach[start], end) # 2. Greedy Iteration (Jump Game II logic) taps = 0 current_end = 0 next_end = 0 # We iterate up to n-1 because if we reach n, the garden is fully watered. for i in range(n): # Update the furthest we can reach from current position i next_end = max(next_end, max_reach[i]) # If we reach the end of the currently active tap's range: if i == current_end: # If we cannot move further than current_end, there is a gap. if next_end <= i: return -1 # "Jump" to the new furthest reach current_end = next_end taps += 1 # Optimization: Early exit if we have already covered the whole garden if current_end >= n: return taps return taps8c:T38f7,

from typing import List

class Solution:
    def minTaps(self, n: int, ranges: List[int]) -> int:
        # Array to store the furthest point reachable from each index.
        # max_reach[i] = max end point for an interval starting at i
        max_reach = [0] * (n + 1)

        # 1. Preprocessing: Convert tap ranges into "jumps"
        for i, r in enumerate(ranges):
            # Calculate the effective bounds of the tap
            start = max(0, i - r)
            end = min(n, i + r)
            
            # We want to know: "If I start at `start`, how far can I go?"
            max_reach[start] = max(max_reach[start], end)

        # 2. Greedy Iteration (Jump Game II logic)
        taps = 0
        current_end = 0
        next_end = 0

        # We iterate up to n-1 because if we reach n, the garden is fully watered.
        for i in range(n):
            # Update the furthest we can reach from current position i
            next_end = max(next_end, max_reach[i])
            
            # If we reach the end of the currently active tap's range:
            if i == current_end:
                # If we cannot move further than current_end, there is a gap.
                if next_end <= i:
                    return -1
                
                # "Jump" to the new furthest reach
                current_end = next_end
                taps += 1
                
                # Optimization: Early exit if we have already covered the whole garden
                if current_end >= n:
                    return taps
                    
        return taps

8d:T261d,

class Solution:
    def videoStitching(self, clips: List[List[int]], time: int) -> int:
        # Sort by start time.
        # End tie-breaking doesn't matter if we take max end during the scan
        clips.sort()

        count = 0
        current_end = 0
        farthest_reach = 0

        i = 0
        n = len(clips)

        # We continue as long as we haven't covered the whole time
        while current_end < time:
            # Process all clips that start before or when the current clip ends
            while i < n and clips[i][0] <= current_end:
                farthest_reach = max(farthest_reach, clips[i][1])
                i += 1
            # If we couldn't find a clip to extend our range, we're stuck!
            if farthest_reach == current_end:
                return -1
            # Let's extend to the farthest possible point!
            current_end = farthest_reach
            count += 1
        return count

8e:T1e79,

def videoStitching(clips, time):
    clips.sort()
    
    # BFS Variables
    current_level_boundary = 0
    next_level_boundary = 0
    jumps = 0
    
    i = 0
    while current_level_boundary < time:
        # Process the "Queue" (all clips start within current range)
        while i < len(clips) and clips[i][0] <= current_level_boundary:
            next_level_boundary = max(next_level_boundary, clips[i][1])
            i += 1
        
        # Dead End Check (Queue empty)
        if current_level_boundary == next_level_boundary:
            return -1
        
        # Move to next level
        current_level_boundary = next_level_boundary
        jumps += 1
        
    return jumps

8f:T5db,class Solution: def maxSumOfThreeSubarrays(self, nums: List[int], k: int) -> List[int]: n = len(nums) # 1) Compute all window sums window_sums = [] curr_sum = sum(nums[:k]) window_sums.append(curr_sum) for i in range(k, n): curr_sum += nums[i] - nums[i - k] window_sums.append(curr_sum) # 2) left[i] = index of best window starting at or before i # Use strict > to keep earliest index (lexicographically smallest) left = [0] * len(window_sums) best = 0 for i in range(len(window_sums)): if window_sums[i] > window_sums[best]: best = i left[i] = best # 3) right[i] = index of best window starting at or after i # Use >= to prefer earlier on ties right = [0] * len(window_sums) best = len(window_sums) - 1 for i in range(len(window_sums) - 1, -1, -1): if window_sums[i] >= window_sums[best]: best = i right[i] = best # 4) Try each middle position max_total = 0 result = 0 for mid in range(k, len(window_sums) - k): l = left[mid - k] # Best left window r = right[mid + k] # Best right window total = window_sums[l] + window_sums[mid] + window_sums[r] if total > max_total: max_total = total result = [l, mid, r] return result90:T591a,

class Solution:
    def maxSumOfThreeSubarrays(self, nums: List[int], k: int) -> List[int]:
        n = len(nums)

        # 1) Compute all window sums
        window_sums = []
        curr_sum = sum(nums[:k])
        window_sums.append(curr_sum)
        for i in range(k, n):
            curr_sum += nums[i] - nums[i - k]
            window_sums.append(curr_sum)
        
        # 2) left[i] = index of best window starting at or before i
        # Use strict > to keep earliest index (lexicographically smallest)
        left = [0] * len(window_sums)
        best = 0
        for i in range(len(window_sums)):
            if window_sums[i] > window_sums[best]:
                best = i
            left[i] = best
        
        # 3) right[i] = index of best window starting at or after i
        # Use >= to prefer earlier on ties
        right = [0] * len(window_sums)
        best = len(window_sums) - 1
        for i in range(len(window_sums) - 1, -1, -1):
            if window_sums[i] >= window_sums[best]:
                best = i
            right[i] = best
        # 4) Try each middle position
        max_total = 0
        result = 0

        for mid in range(k, len(window_sums) - k):
            l = left[mid - k] # Best left window
            r = right[mid + k] # Best right window
            total = window_sums[l] + window_sums[mid] + window_sums[r]

            if total > max_total:
                max_total = total
                result = [l, mid, r]
        return result

91:T588,class Solution: def findLengthOfShortestSubarray(self, arr: List[int]) -> int: n = len(arr) # 1. Find the longest non-decreasing prefix (left portion) left = 0 while left < n - 1 and arr[left] <= arr[left + 1]: left += 1 # Postcondition: left marks the inclusive end of the leftmost sorted portion # If the whole array is sorted: if left == n - 1: return 0 # 2. Find the longest non-decreasing suffix (right portion) right = n - 1 while right > 0 and arr[right - 1] <= arr[right]: right -= 1 # 3. Initial answer: remove either all prefix or all suffix # Removing everything after arr[left] (length = n - 1 - left) # Removing everything before arr[right] (length = right) result = min(n - left - 1, right) # 4. Use two pointers to merge prefix and suffix i = 0 j = right while i <= left and j < n: if arr[i] <= arr[j]: # Valid merge: we keep arr[0..i] and arr[j..n-1] # Elements removed are between i and j # Count = j - i - 1 result = min(result, j - i - 1) i += 1 else: # arr[i] is too big; we need a larger element from the right side j += 1 return result 92:T3a2d,

class Solution:
    def findLengthOfShortestSubarray(self, arr: List[int]) -> int:
        n = len(arr)

        # 1. Find the longest non-decreasing prefix (left portion)
        left = 0
        while left < n - 1 and arr[left] <= arr[left + 1]:
            left += 1
        # Postcondition: left marks the inclusive end of the leftmost sorted portion

        # If the whole array is sorted:
        if left == n - 1:
            return 0
        
        # 2. Find the longest non-decreasing suffix (right portion)
        right = n - 1
        while right > 0 and arr[right - 1] <= arr[right]:
            right -= 1
        
        # 3. Initial answer: remove either all prefix or all suffix
        # Removing everything after arr[left] (length = n - 1 - left)
        # Removing everything before arr[right] (length = right)
        result = min(n - left - 1, right)

        # 4. Use two pointers to merge prefix and suffix
        i = 0
        j = right
        
        while i <= left and j < n:
            if arr[i] <= arr[j]:
                # Valid merge: we keep arr[0..i] and arr[j..n-1]
                # Elements removed are between i and j
                # Count = j - i - 1
                result = min(result, j - i - 1)
                i += 1
            else:
                # arr[i] is too big; we need a larger element from the right side
                j += 1
        return result

93:T5bb,class Solution: def recoverTree(self, root: Optional[TreeNode]) -> None: # Variables to keep track of the two nodes to be swapped self.first = None self.second = None # Variable to keep track of the previous node in the In-Order traversal # We initialize it with a node having -infinity to handle the first comparison safely self.prev = TreeNode(float('-inf')) def inorder(node): if not node: return # 1. Visit left inorder(node.left) # 2. Visit Node (the check) # If the sorting is broken (prev is greater than current) if self.prev.val > node.val: # If it's the first time we've seen a violation: if not self.first: self.first = self.prev # The larget value is the problem self.second = node # The smaller value is the problem (tentatively) else: # If this is the second time we've seen a violation: self.second = node # The new smaller value is the actual second node # Update prev to current before moving on self.prev = node # 3. Visit Right inorder(node.right) # Run the traversal inorder(root) # Swap the values self.first.val, self.second.val = self.second.val, self.first.val94:T380b,

class Solution:
    def recoverTree(self, root: Optional[TreeNode]) -> None:
        # Variables to keep track of the two nodes to be swapped
        self.first = None
        self.second = None

        # Variable to keep track of the previous node in the In-Order traversal
        # We initialize it with a node having -infinity to handle the first comparison safely
        self.prev = TreeNode(float('-inf'))

        def inorder(node):
            if not node:
                return
            
            # 1. Visit left
            inorder(node.left)

            # 2. Visit Node (the check)
            # If the sorting is broken (prev is greater than current)
            if self.prev.val > node.val:
                # If it's the first time we've seen a violation:
                if not self.first:
                    self.first = self.prev # The larget value is the problem
                    self.second = node # The smaller value is the problem (tentatively)
                else:
                    # If this is the second time we've seen a violation:
                    self.second = node # The new smaller value is the actual second node
            # Update prev to current before moving on
            self.prev = node

            # 3. Visit Right
            inorder(node.right)
        
        # Run the traversal
        inorder(root)

        # Swap the values
        self.first.val, self.second.val = self.second.val, self.first.val

95:T45b,class Solution: def recoverTree(self, root: Optional[TreeNode]) -> None: stack = [] node = root prev = None # Tracks the previously visited node in-order first = None # The first node to be swapped second = None # The second node to be swapped while stack or node: # 1. Drill left (find the smallest remaining value) while node: stack.append(node) node = node.left # 2. Process the node (this is the "In-Order" part) node = stack.pop() # CHECK FOR VIOLATION if prev and prev.val > node.val: if not first: first = prev second = node # Tentative else: second = node # Found the second drop # Update prev prev = node # 3. Go Right node = node.right # 4. Perform the single swap first.val, second.val = second.val, first.val96:T2c07,

class Solution:
    def recoverTree(self, root: Optional[TreeNode]) -> None:
        stack = []
        node = root
        
        prev = None # Tracks the previously visited node in-order
        first = None # The first node to be swapped
        second = None # The second node to be swapped
        
        while stack or node:
            # 1. Drill left (find the smallest remaining value)
            while node:
                stack.append(node)
                node = node.left
            
            # 2. Process the node (this is the "In-Order" part)
            node = stack.pop()
            
            # CHECK FOR VIOLATION
            if prev and prev.val > node.val:
                if not first:
                    first = prev
                    second = node # Tentative
                else:
                    second = node # Found the second drop
            
            # Update prev
            prev = node
            
            # 3. Go Right
            node = node.right
            
        # 4. Perform the single swap
        first.val, second.val = second.val, first.val

97:T5fc,class Solution: def shortestPathLength(self, graph: List[List[int]]) -> int: n = len(graph) # Edge case: If there's only 1 node, path length is 0 if n == 1: return 0 # The final state we want (all bits set to 1) # Example: if n = 3, ending_mask = 111 (binary) = 7 ending_mask = (1 << n) - 1 # Queue for BFS: (current_node, current_mask, distance) queue = deque() # Set to keep track of visited states: (current_node, current_mask) # This prevents us from doing redundant work or getting stuck in loops seen = set() # Initialize BFS from all nodes # We can start anywhere, so we treat all nodes as possible starting points for i in range(n): initial_mask = 1 << i queue.append((i, initial_mask, 0)) seen.add((i, initial_mask)) while queue: node, mask, dist = queue.popleft() # Check if we've visited all nodes if mask == ending_mask: return dist # Traverse neighbors for neighbor in graph[node]: new_mask = mask | (1 << neighbor) # If we haven't been at this neighbor with this specific path history if (neighbor, new_mask) not in seen: seen.add((neighbor, new_mask)) queue.append((neighbor, new_mask, dist + 1)) return -1 # If the graph is connected, this won't happen98:T39bb,

class Solution:
    def shortestPathLength(self, graph: List[List[int]]) -> int:
        n = len(graph)

        # Edge case: If there's only 1 node, path length is 0
        if n == 1:
            return 0

        # The final state we want (all bits set to 1)
        # Example: if n = 3, ending_mask = 111 (binary) = 7
        ending_mask = (1 << n) - 1

        # Queue for BFS: (current_node, current_mask, distance)
        queue = deque()

        # Set to keep track of visited states: (current_node, current_mask)
        # This prevents us from doing redundant work or getting stuck in loops
        seen = set()
        # Initialize BFS from all nodes
        # We can start anywhere, so we treat all nodes as possible starting points
        for i in range(n):
            initial_mask = 1 << i
            queue.append((i, initial_mask, 0))
            seen.add((i, initial_mask))

        while queue:
            node, mask, dist = queue.popleft()

            # Check if we've visited all nodes
            if mask == ending_mask:
                return dist
            
            # Traverse neighbors
            for neighbor in graph[node]:
                new_mask = mask | (1 << neighbor)

                # If we haven't been at this neighbor with this specific path history
                if (neighbor, new_mask) not in seen:
                    seen.add((neighbor, new_mask))
                    queue.append((neighbor, new_mask, dist + 1))
        return -1 # If the graph is connected, this won't happen

99:T744,The fundamental secret of Monotonic Stacks: The stack always stores the "unresolved" problems, and we pop when we find the element that "resolves" them. In every stack problem, you just have to identify "What is the event that stops the waiting game?" If the event is "finding a bigger number," you keep a decreasing stack. If the event is "finding a smaller number," you keep an increasing stack. There is a spiritual link I feel here between monotonic stack problems and the optimal Longest Increasing Subsequence solution. That "spirit" I'm detecting is the concept of maintaining a candidate frontier. Both the Monotonic Stack and the LIS (Patience Sorting/Tails) algorithms belong to a specific family of greedy algorithms that work by discarding suboptimal possibilities to maintain a sorted invariant. The Core Philosophy: "Survival of the Fittest" Both algorithms rely on the fact that if the data structure isn't sorted, we are holding onto redundant information. The Difference: Replacement vs. Resolution This is where they diverge, and why they solve different problems. It comes down to what you do when we break the sort order. Scenario: The structure contains [... 10] and the new number is 5. The LIS Approach (Replacement): 5 is better than 10 for that position. Action: Overwrite. [... 10] becomes [... 5]. Meaning: "I found a better way to reach this state." Scenario: The structure contains [... 5] and the new number is 10. The Mono Stack Approach (Resolution): 10 solves the problem for 5. Action: Pop 5. Meaning: "The 5 is done. It found what it was looking for. Now let's see if 10 solves the guy behind 5." We can think of both algorithms as maintaining a Pareto Frontier of valid states. They both iterate and greedily maintain a sorted structure of "active candidates" and discarding anything that is no longer useful.9a:T7c8,

The fundamental secret of Monotonic Stacks:

The stack always stores the "unresolved" problems, and we pop when we find the element that "resolves" them.

In every stack problem, you just have to identify "What is the event that stops the waiting game?"

If the event is "finding a bigger number," you keep a decreasing stack.

If the event is "finding a smaller number," you keep an increasing stack.

There is a spiritual link I feel here between monotonic stack problems and the optimal Longest Increasing Subsequence solution. That "spirit" I'm detecting is the concept of maintaining a candidate frontier.

Both the Monotonic Stack and the LIS (Patience Sorting/Tails) algorithms belong to a specific family of greedy algorithms that work by discarding suboptimal possibilities to maintain a sorted invariant.

The Core Philosophy: "Survival of the Fittest"

Both algorithms rely on the fact that if the data structure isn't sorted, we are holding onto redundant information.

The Difference: Replacement vs. Resolution

This is where they diverge, and why they solve different problems. It comes down to what you do when we break the sort order.

Scenario: The structure contains [... 10] and the new number is 5.
The LIS Approach (Replacement):
5 is better than 10 for that position.
Action: Overwrite. [... 10] becomes [... 5].
Meaning: "I found a better way to reach this state."

Scenario: The structure contains [... 5] and the new number is 10.
The Mono Stack Approach (Resolution):
10 solves the problem for 5.
Action: Pop 5.
Meaning: "The 5 is done. It found what it was looking for. Now let's see if 10 solves the guy behind 5."

We can think of both algorithms as maintaining a Pareto Frontier of valid states.

They both iterate and greedily maintain a sorted structure of "active candidates" and discarding anything that is no longer useful.

9b:T214a,

class Solution:
    def nextGreaterElement(self, nums1, nums2):
        # Dictionary to store the next greater element for each number in nums2
        next_greater_map = {}
        stack = []
        
        # Iterate through nums2 to build the map
        for num in nums2:
            # While stack is not empty and current number is greater than stack top
            while stack and num > stack[-1]:
                val = stack.pop()
                next_greater_map[val] = num
            stack.append(num)
        
        # Iterate through nums1 to build the result
        # If the number exists in map, use the value. Otherwise -1.
        res = []
        for num in nums1:
            if num in next_greater_map:
                res.append(next_greater_map[num])
            else:
                res.append(-1)
                
        return res

9c:T52c,Parentheses simply denote order, so we can just try everything. We iterate through every possible pair at every stage. Time Complexity Analysis At first glance, this looks expensive, but for N=4, it is very small. This is technically O(1) because the input size is fixed, but let's derive the generalized complexity to show the interviewer we know our math. We have 3 stages of recursion for 4 numbers. Stage 1: 4 numbers left We choose an ordered pair of numbers (i, j). There are 4 choices for i and 3 choices for j (since i≠j). Total pairs: 4×3=12 pairs. For each pair, we run 4 operations (+, -, *, /). Total branches: 12×4=48. Stage 2: 3 numbers left We passed 1 result + 2 existing numbers = 3 numbers. We choose an ordered pair.3×2=6 pairs. 4 operations. Total branches: 6×4=24. Stage 3: 2 numbers left We passed 1 result + 1 existing number = 2 numbers. We choose an ordered pair. 2×1=2 pairs. 4 operations. Total branches: 2×4=8 Total Operations (Upper Bound): 48×24×8=9,216 We perform roughly 9,216 recursive calls in the worst case (if we find no solution). For a computer, 9,000 operations is instantaneous. Note: The actual number is slightly lower because we don't divide by zero, and we stop as soon as we find True. https://aistudio.google.com/prompts/1QfU4hBSamwdtDapaQjDvDeQZA-smhQ8w9d:T615,

Parentheses simply denote order, so we can just try everything. We iterate through every possible pair at every stage.

Time Complexity Analysis
At first glance, this looks expensive, but for N=4, it is very small. This is technically O(1) because the input size is fixed, but let's derive the generalized complexity to show the interviewer we know our math.

We have 3 stages of recursion for 4 numbers.

Stage 1: 4 numbers left

We choose an ordered pair of numbers (i, j).
There are 4 choices for i and 3 choices for j (since i≠j).

Total pairs: 4×3=12 pairs.
For each pair, we run 4 operations (+, -, *, /).
Total branches: 12×4=48.

Stage 2: 3 numbers left
We passed 1 result + 2 existing numbers = 3 numbers.
We choose an ordered pair.3×2=6 pairs.
4 operations.

Total branches: 6×4=24.

Stage 3: 2 numbers left
We passed 1 result + 1 existing number = 2 numbers.
We choose an ordered pair. 2×1=2 pairs.
4 operations.

Total branches:
2×4=8

Total Operations (Upper Bound):
48×24×8=9,216

We perform roughly 9,216 recursive calls in the worst case (if we find no solution).
For a computer, 9,000 operations is instantaneous.
Note: The actual number is slightly lower because we don't divide by zero, and we stop as soon as we find True.

https://aistudio.google.com/prompts/1QfU4hBSamwdtDapaQjDvDeQZA-smhQ8w

9e:T9a4,class Solution: def judgePoint24(self, cards: List[int]) -> bool: # Use a small epsilon for float comparison EPSILON = 1e-6 def backtrack(nums): # Base Case: If only one number remains, check if it's 24 if len(nums) == 1: return abs(nums[0] - 24) < EPSILON # Try picking every possible pair of numbers (i, j) for i in range(len(nums)): for j in range(len(nums)): # Must pick two different cards if i == j: continue # Create a list of the numbers NOT picked next_nums = [] for k in range(len(nums)): if k != i and k != j: next_nums.append(nums[k]) # We will try all operations on nums[i] and nums[j] # and append the result to next_nums val1 = nums[i] val2 = nums[j] # Possible results from operations # Note: For + and *, order doesn't matter, but our loop # tries (i,j) and (j,i), so we technically compute it twice. # That is fine for n=4. # 1. Addition next_nums.append(val1 + val2) if backtrack(next_nums): return True next_nums.pop() # Backtrack # 2. Multiplication next_nums.append(val1 * val2) if backtrack(next_nums): return True next_nums.pop() # 3. Subtraction (val1 - val2) # The loop will eventually swap i and j to cover (val2 - val1) next_nums.append(val1 - val2) if backtrack(next_nums): return True next_nums.pop() # 4. Division (val1 / val2) if abs(val2) > EPSILON: # Avoid division by zero next_nums.append(val1 / val2) if backtrack(next_nums): return True next_nums.pop() return False return backtrack(cards)9f:T5518,

class Solution:
    def judgePoint24(self, cards: List[int]) -> bool:
        # Use a small epsilon for float comparison
        EPSILON = 1e-6
        
        def backtrack(nums):
            # Base Case: If only one number remains, check if it's 24
            if len(nums) == 1:
                return abs(nums[0] - 24) < EPSILON
            
            # Try picking every possible pair of numbers (i, j)
            for i in range(len(nums)):
                for j in range(len(nums)):
                    # Must pick two different cards
                    if i == j:
                        continue
                    
                    # Create a list of the numbers NOT picked
                    next_nums = []
                    for k in range(len(nums)):
                        if k != i and k != j:
                            next_nums.append(nums[k])
                    
                    # We will try all operations on nums[i] and nums[j]
                    # and append the result to next_nums
                    
                    val1 = nums[i]
                    val2 = nums[j]
                    
                    # Possible results from operations
                    # Note: For + and *, order doesn't matter, but our loop 
                    # tries (i,j) and (j,i), so we technically compute it twice.
                    # That is fine for n=4.
                    
                    # 1. Addition
                    next_nums.append(val1 + val2)
                    if backtrack(next_nums): return True
                    next_nums.pop() # Backtrack
                    
                    # 2. Multiplication
                    next_nums.append(val1 * val2)
                    if backtrack(next_nums): return True
                    next_nums.pop()
                    
                    # 3. Subtraction (val1 - val2)
                    # The loop will eventually swap i and j to cover (val2 - val1)
                    next_nums.append(val1 - val2)
                    if backtrack(next_nums): return True
                    next_nums.pop()
                    
                    # 4. Division (val1 / val2)
                    if abs(val2) > EPSILON: # Avoid division by zero
                        next_nums.append(val1 / val2)
                        if backtrack(next_nums): return True
                        next_nums.pop()
            
            return False

        return backtrack(cards)

a0:T66c,1. The Trigger: Why do we only calculate when heights drop? This is the most critical intuition. Imagine you are standing at index i looking left. You see a building of height H. You want to know: "How far to the right can this rectangle extend?" If the next building is taller: That's great! The current rectangle of height H can continue through the taller building. We don't need to stop. We don't know the limit yet. So, we just push it onto the stack and wait. If the next building is shorter: This is the "bad news." The rectangle of height H cannot pass through a shorter building. Its run is over. This "drop" in height is the Right Boundary for the tall bar we just passed. This is the only moment we have enough information to calculate the area for that tall bar, so we pop it and do the math. 2. The Width Formula: i - stack[-1] - 1 This formula is calculating the distance between the "Right Wall" and the "Left Wall" that contain the popped bar. Let's look at the specific moment we pop a bar. The Bar we act on: Let's say we just popped a bar at index j. Its height is H. The Right Boundary (i): We are currently at index i. We triggered the pop because heights[i] is shorter than H. So i is the first index to the right where the rectangle cannot exist. (Exclusive Right Bound) The Left Boundary (stack[-1]): After popping j, the new top of the stack is stack[-1]. This index represents the first bar to the left that was shorter than H. (Exclusive Left Bound) Why -1? We want the number of bars between stack[-1] and i. Math: If you have boundaries L and R (exclusive), the count of items between them is R - L - 1.a1:T6da,

The Trigger: Why do we only calculate when heights drop?
This is the most critical intuition.

Imagine you are standing at index i looking left. You see a building of height H. You want to know: "How far to the right can this rectangle extend?"

If the next building is taller: That's great! The current rectangle of height H can continue through the taller building. We don't need to stop. We don't know the limit yet. So, we just push it onto the stack and wait.

If the next building is shorter: This is the "bad news." The rectangle of height H cannot pass through a shorter building. Its run is over.

This "drop" in height is the Right Boundary for the tall bar we just passed.

This is the only moment we have enough information to calculate the area for that tall bar, so we pop it and do the math.

The Width Formula: i - stack[-1] - 1
This formula is calculating the distance between the "Right Wall" and the "Left Wall" that contain the popped bar.

Let's look at the specific moment we pop a bar.

The Bar we act on: Let's say we just popped a bar at index j. Its height is H.

The Right Boundary (i): We are currently at index i. We triggered the pop because heights[i] is shorter than H. So i is the first index to the right where the rectangle cannot exist. (Exclusive Right Bound)

The Left Boundary (stack[-1]): After popping j, the new top of the stack is stack[-1]. This index represents the first bar to the left that was shorter than H. (Exclusive Left Bound)

Why -1? We want the number of bars between stack[-1] and i.

Math: If you have boundaries L and R (exclusive), the count of items between them is R - L - 1.

a2:T655,class Solution: def maximalRectangle(self, matrix: list[list[str]]) -> int: if not matrix: return 0 rows, cols = len(matrix), len(matrix[0]) heights = [0] * cols max_area = 0 for r in range(rows): for c in range(cols): # Update the histogram heights if matrix[r][c] == "1": heights[c] += 1 else: heights[c] = 0 # Calculate max area for this row's histogram max_area = max(max_area, self.largestRectangleArea(heights)) return max_area # Helper: Solves the 1D Histogram problem def largestRectangleArea(self, heights: list[int]) -> int: stack = [] # stores indices max_a = 0 # Append a 0 height at the end to force pop remaining bars # We perform a shallow copy + append to avoid modifying the original heights array repeatedly current_heights = heights + [0] for i, h in enumerate(current_heights): start = i # While the current bar is shorter than the bar at stack top, # pop the stack and calculate area. while stack and current_heights[stack[-1]] > h: index = stack.pop() height = current_heights[index] # Width is current index - index of the previous item in stack - 1 width = i if not stack else i - stack[-1] - 1 max_a = max(max_a, height * width) stack.append(i) return max_aa3:T4c6d,

class Solution:
    def maximalRectangle(self, matrix: list[list[str]]) -> int:
        if not matrix:
            return 0

        rows, cols = len(matrix), len(matrix[0])
        heights = [0] * cols
        max_area = 0

        for r in range(rows):
            for c in range(cols):
                # Update the histogram heights
                if matrix[r][c] == "1":
                    heights[c] += 1
                else:
                    heights[c] = 0
            
            # Calculate max area for this row's histogram
            max_area = max(max_area, self.largestRectangleArea(heights))
        
        return max_area

    # Helper: Solves the 1D Histogram problem
    def largestRectangleArea(self, heights: list[int]) -> int:
        stack = [] # stores indices
        max_a = 0
        # Append a 0 height at the end to force pop remaining bars
        # We perform a shallow copy + append to avoid modifying the original heights array repeatedly
        current_heights = heights + [0]
        
        for i, h in enumerate(current_heights):
            start = i
            # While the current bar is shorter than the bar at stack top,
            # pop the stack and calculate area.
            while stack and current_heights[stack[-1]] > h:
                index = stack.pop()
                height = current_heights[index]
                # Width is current index - index of the previous item in stack - 1
                width = i if not stack else i - stack[-1] - 1
                max_a = max(max_a, height * width)
            stack.append(i)
            
        return max_a

a4:T83e,class Solution: def solveSudoku(self, board: List[List[str]]) -> None: # 1. State Management (The "Sets") # Use simple arrays/sets to track what numbers are currently in use rows = [set() for _ in range(9)] cols = [set() for _ in range(9)] boxes = [set() for _ in range(9)] # 2. Pre-process the board # Fill our sets with the initial numbers given in the puzzle for r in range(9): for c in range(9): if board[r][c] != ".": val = board[r][c] rows[r].add(val) cols[c].add(val) boxes[(r // 3) * 3 + (c // 3)].add(val) # 3. The Backtracking Function def backtrack(r, c): # If we went past the last column, move to next row if c == 9: return backtrack(r + 1, 0) # If we went past the last row, we are done! Solution found. if r == 9: return True # If cell is already filled, skip it if board[r][c] != ".": return backtrack(r, c + 1) # Identify the 3x3 box index box_idx = (r // 3) * 3 + (c // 3) # Try numbers 1-9 for k in range(1, 10): val = str(k) # CHECK CONSTRAINTS (O(1) lookup) if val not in rows[r] and val not in cols[c] and val not in boxes[box_idx]: # PLACE board[r][c] = val rows[r].add(val) cols[c].add(val) boxes[box_idx].add(val) # RECURSE if backtrack(r, c + 1): return True # BACKTRACK (Remove) board[r][c] = "." rows[r].remove(val) cols[c].remove(val) boxes[box_idx].remove(val) return False backtrack(0, 0)a5:T728b,

class Solution:
    def solveSudoku(self, board: List[List[str]]) -> None:
        # 1. State Management (The "Sets")
        # Use simple arrays/sets to track what numbers are currently in use
        rows = [set() for _ in range(9)]
        cols = [set() for _ in range(9)]
        boxes = [set() for _ in range(9)]

        # 2. Pre-process the board
        # Fill our sets with the initial numbers given in the puzzle
        for r in range(9):
            for c in range(9):
                if board[r][c] != ".":
                    val = board[r][c]
                    rows[r].add(val)
                    cols[c].add(val)
                    boxes[(r // 3) * 3 + (c // 3)].add(val)

        # 3. The Backtracking Function
        def backtrack(r, c):
            # If we went past the last column, move to next row
            if c == 9:
                return backtrack(r + 1, 0)
            
            # If we went past the last row, we are done! Solution found.
            if r == 9:
                return True
            
            # If cell is already filled, skip it
            if board[r][c] != ".":
                return backtrack(r, c + 1)
            
            # Identify the 3x3 box index
            box_idx = (r // 3) * 3 + (c // 3)

            # Try numbers 1-9
            for k in range(1, 10):
                val = str(k)
                # CHECK CONSTRAINTS (O(1) lookup)
                if val not in rows[r] and val not in cols[c] and val not in boxes[box_idx]:
                    
                    # PLACE
                    board[r][c] = val
                    rows[r].add(val)
                    cols[c].add(val)
                    boxes[box_idx].add(val)

                    # RECURSE
                    if backtrack(r, c + 1):
                        return True
                    
                    # BACKTRACK (Remove)
                    board[r][c] = "."
                    rows[r].remove(val)
                    cols[c].remove(val)
                    boxes[box_idx].remove(val)
            
            return False

        backtrack(0, 0)

a6:T1cb1,

class Solution:
    def findDisappearedNumbers(self, nums):
        # Pass 1: Mark existing numbers by negating values at the corresponding index
        for n in nums:
            # We use abs() because the number might have been negated already by a previous step
            idx = abs(n) - 1
            
            # Only negate if it's currently positive
            if nums[idx] > 0:
                nums[idx] *= -1

        # Pass 2: Collect indices that contain positive numbers
        res = []
        for i in range(len(nums)):
            if nums[i] > 0:
                # If the number at index i is positive, i+1 was missing
                res.append(i + 1)
                
        return res

a7:T232f,

class Solution:
    def searchMatrix(self, matrix, target):
        if not matrix or not matrix[0]:
            return False
        
        m, n = len(matrix), len(matrix[0])
        
        # Start at Top-Right Corner
        row = 0
        col = n - 1
        
        # Keep going as long as we are inside the matrix boundaries
        while row < m and col >= 0:
            current_val = matrix[row][col]
            
            if current_val == target:
                return True
            
            elif current_val > target:
                # Too big? Eliminate this column, move left
                col -= 1
                
            else: # current_val < target
                # Too small? Eliminate this row, move down
                row += 1
                
        return False

a8:T2748,

class Solution:
    def isPalindrome(self, head):
        if not head or not head.next:
            return True

        # 1. Find the Middle using Fast/Slow pointers
        slow = head
        fast = head
        while fast and fast.next:
            slow = slow.next
            fast = fast.next.next
        
        # 2. Reverse the second half in-place
        # 'prev' will eventually become the head of the reversed half
        prev = None
        curr = slow # 'slow' is roughly the middle
        while curr:
            next_temp = curr.next
            curr.next = prev
            prev = curr
            curr = next_temp
            
        # 3. Compare the two halves
        left = head
        right = prev # 'prev' is now the start of the reversed right half
        
        while right: # We only check until the right half ends
            if left.val != right.val:
                return False
            left = left.next
            right = right.next
            
        return True

a9:Tfc6,

class Solution:
    def isPalindrome(self, head):
        vals = []
        current = head
        # 1. Copy to list
        while current:
            vals.append(current.val)
            current = current.next
        
        # 2. Check if list is palindrome
        return vals == vals[::-1]

aa:Teab,

# 1. READ (Safe)
# We are just looking at the map and remembering where to go next.
# No structure changes.
next_temp = curr.next 

# 2. MUTATE (The "One Thing")
# This is the ONLY time we touch the heap/memory structure.
# We flip exactly one arrow.
curr.next = prev

# 3. MOVE (Safe)
# We just update our own internal label 'prev' to point to the current node.
# No structure changes.
prev = curr

# 4. MOVE (Safe)
# We update our label 'curr' to point to the saved location.
# No structure changes.
curr = next_temp

ab:T471,class Solution: def zigzagLevelOrder(self, root): if not root: return [] results = [] queue = deque([root]) left_to_right = True # The toggle flag while queue: level_size = len(queue) current_level_values = deque() # Use deque for O(1) front insertions for _ in range(level_size): node = queue.popleft() # PRESENTATION LOGIC: Handle the zigzag if left_to_right: current_level_values.append(node.val) else: current_level_values.appendleft(node.val) # TRAVERSAL LOGIC: Always standard L -> R if node.left: queue.append(node.left) if node.right: queue.append(node.right) # Add the current level to results and toggle the flag results.append(list(current_level_values)) left_to_right = not left_to_right return resultsac:T2b49,

class Solution:
    def zigzagLevelOrder(self, root):
        if not root:
            return []
        
        results = []
        queue = deque([root])
        left_to_right = True # The toggle flag
        
        while queue:
            level_size = len(queue)
            current_level_values = deque() # Use deque for O(1) front insertions
            
            for _ in range(level_size):
                node = queue.popleft()
                
                # PRESENTATION LOGIC: Handle the zigzag
                if left_to_right:
                    current_level_values.append(node.val)
                else:
                    current_level_values.appendleft(node.val)
                    
                # TRAVERSAL LOGIC: Always standard L -> R
                if node.left:
                    queue.append(node.left)
                if node.right:
                    queue.append(node.right)
            
            # Add the current level to results and toggle the flag
            results.append(list(current_level_values))
            left_to_right = not left_to_right
            
        return results

ad:T1ad7,

class Solution:
    def shortestPalindrome(self, s: str) -> str:
        # 1. Reverse the string
        r = s[::-1]
        
        # 2. Iterate through the length of the string
        # We want to find the largest 'i' such that 
        # the prefix of s matches the suffix of r
        for i in range(len(s) + 1):
            # s.startswith(r[i:]) checks if the beginning of s 
            # matches the reversed string minus the first i characters.
            if s.startswith(r[i:]):
                # The part that didn't match (r[:i]) is exactly 
                # the reverse of the suffix we need to add.
                return r[:i] + s
                
        return "" # Should basically never hit here given s starts with s

ae:T518,def shortestPalindrome_rolling_hash(s: str) -> str: n = len(s) if n == 0: return "" # Constants for hashing base = 29 mod = 10**9 + 7 forward_hash = 0 reverse_hash = 0 power_value = 1 palindrome_end_index = -1 for i, char in enumerate(s): val = ord(char) - ord('a') + 1 # 1. Update Forward Hash: # Think of this as adding a term to a polynomial: # Hash("abc") = a * base^0 + b * base^1 + c * base^2 forward_hash = (forward_hash + val * power_value) % mod # 2. Update Reverse Hash: # Think of this as shifting the existing hash and adding the new char at the end: # Step 1 "a": a # Step 2 "ab": a * base + b # Step 3 "abc": (a * base + b) * base + c reverse_hash = (reverse_hash * base + val) % mod # 3. Check for Palindrome if forward_hash == reverse_hash: palindrome_end_index = i # Update power for the next forward_hash iteration power_value = (power_value * base) % mod # We found the longest palindromic prefix ends at palindrome_end_index # The suffix is everything after that suffix = s[palindrome_end_index + 1:] return suffix[::-1] + saf:T349e,

def shortestPalindrome_rolling_hash(s: str) -> str:
    n = len(s)
    if n == 0: return ""
    
    # Constants for hashing
    base = 29
    mod = 10**9 + 7
    
    forward_hash = 0
    reverse_hash = 0
    power_value = 1
    palindrome_end_index = -1
    
    for i, char in enumerate(s):
        val = ord(char) - ord('a') + 1
        
        # 1. Update Forward Hash:
        # Think of this as adding a term to a polynomial:
        # Hash("abc") = a * base^0 + b * base^1 + c * base^2
        forward_hash = (forward_hash + val * power_value) % mod
        
        # 2. Update Reverse Hash:
        # Think of this as shifting the existing hash and adding the new char at the end:
        # Step 1 "a":   a
        # Step 2 "ab":  a * base + b
        # Step 3 "abc": (a * base + b) * base + c
        reverse_hash = (reverse_hash * base + val) % mod
        
        # 3. Check for Palindrome
        if forward_hash == reverse_hash:
            palindrome_end_index = i
            
        # Update power for the next forward_hash iteration
        power_value = (power_value * base) % mod
        
    # We found the longest palindromic prefix ends at palindrome_end_index
    # The suffix is everything after that
    suffix = s[palindrome_end_index + 1:]
    
    return suffix[::-1] + s

b0:T721,class Solution: def shortestBridge(self, grid: list[list[int]]) -> int: n = len(grid) directions = [(0, 1), (0, -1), (1, 0), (-1, 0)] visit = set() q = deque() # Helper: DFS to find the first island completely def dfs(r, c): if (r < 0 or c < 0 or r >= n or c >= n or grid[r][c] == 0 or (r, c) in visit): return visit.add((r, c)) q.append((r, c)) # Add to queue for BFS later for dr, dc in directions: dfs(r + dr, c + dc) # 1. FIND THE FIRST ISLAND found = False for r in range(n): if found: break for c in range(n): if grid[r][c] == 1: dfs(r, c) found = True break # 2. BFS TOWARD THE SECOND ISLAND res = 0 while q: # Process entire layer (level-by-level) for _ in range(len(q)): r, c = q.popleft() for dr, dc in directions: nr, nc = r + dr, c + dc if 0 <= nr < n and 0 <= nc < n: if (nr, nc) in visit: continue if grid[nr][nc] == 1: return res # Found the second island! # Use the set to mark water as visited so we don't loop visit.add((nr, nc)) q.append((nr, nc)) # Increment distance after finishing a full layer of expansion res += 1 return -1b1:T5ed0,

class Solution:
    def shortestBridge(self, grid: list[list[int]]) -> int:
        n = len(grid)
        directions = [(0, 1), (0, -1), (1, 0), (-1, 0)]
        
        visit = set()
        q = deque()
        
        # Helper: DFS to find the first island completely
        def dfs(r, c):
            if (r < 0 or c < 0 or r >= n or c >= n or 
                grid[r][c] == 0 or (r, c) in visit):
                return
            
            visit.add((r, c))
            q.append((r, c)) # Add to queue for BFS later
            
            for dr, dc in directions:
                dfs(r + dr, c + dc)
        
        # 1. FIND THE FIRST ISLAND
        found = False
        for r in range(n):
            if found: break
            for c in range(n):
                if grid[r][c] == 1:
                    dfs(r, c)
                    found = True
                    break
        
        # 2. BFS TOWARD THE SECOND ISLAND
        res = 0
        while q:
            # Process entire layer (level-by-level)
            for _ in range(len(q)):
                r, c = q.popleft()
                
                for dr, dc in directions:
                    nr, nc = r + dr, c + dc
                    
                    if 0 <= nr < n and 0 <= nc < n:
                        if (nr, nc) in visit:
                            continue
                        
                        if grid[nr][nc] == 1:
                            return res # Found the second island!
                        
                        # Use the set to mark water as visited so we don't loop
                        visit.add((nr, nc))
                        q.append((nr, nc))
            
            # Increment distance after finishing a full layer of expansion
            res += 1
            
        return -1

b2:T9b4,

class Solution:
    def reverseWords(self, s: str) -> str:
        return ' '.join(s.split()[::-1])

b3:T2dad,

class Solution:
    def reverseWords(self, s: str) -> str:
        left, right = 0, len(s) - 1
        
        # Trim edges first to reduce loop iterations
        while left <= right and s[left] == ' ':
            left += 1
        while left <= right and s[right] == ' ':
            right -= 1
            
        queue = deque()
        word = []
        
        while left <= right:
            if s[left] != ' ':
                # Build the current word char by char
                word.append(s[left])
            elif word:
                # We hit a space! Push the complete word to the FRONT
                queue.appendleft("".join(word))
                word = []
            
            left += 1
            
        # Don't forget the last word (since the string doesn't end with a space)
        if word:
            queue.appendleft("".join(word))
            
        return " ".join(queue)

b4:T295b,

class Solution:
    def getHint(self, secret: str, guess: str) -> str:
        bulls = 0
        
        # We will store the characters that are NOT bulls here
        secret_leftovers = []
        guess_leftovers = []
        
        # Pass 1: Find Bulls and filter out non-matches
        for s, g in zip(secret, guess):
            if s == g:
                bulls += 1
            else:
                secret_leftovers.append(s)
                guess_leftovers.append(g)
        
        # Pass 2: Count Cows using the leftovers
        # We count the frequency of digits in both leftover lists
        s_counts = Counter(secret_leftovers)
        g_counts = Counter(guess_leftovers)
        
        cows = 0
        # The number of cows for a specific digit is the minimum 
        # of its occurrence in secret vs guess.
        for digit in s_counts:
            cows += min(s_counts[digit], g_counts[digit])
            
        return f"{bulls}A{cows}B"

b5:T48f,

We make a lookup table that's essentially a set of instructions for our "read pointer."

We store a specific tuple (length_of_source, target) at indices in order to control the flow of the read pointer i.

Why the key is index:

As we iterate through the original string s (from 0 to len(s)), we need to ask at every single step: "Do I behave normally here, or do I do something special?"
The hash map (dictionary) allows us to answer that question in O(1) (instant) time. Without it, we'd have to scan the indices array constantly.

Why Store length_of_source (The Skip):

Since we are not mutating the string in place, we effectively "delete" the old substring by simply ignoring it.
If we find a match at index i, we need to know exactly how many characters to "jump" over so we don't accidentally copy parts of the old string we intended to replace.

Why Store target (The Write):

This is just the payload. "Since I'm skipping the old stuff, what do I put in the result list instead?"

b6:T455,class Solution: def findReplaceString(self, s: str, indices: List[int], sources: List[str], targets: List[str]) -> str: # Map to store valid replacements: index -> (length_of_source, targets) lookup = {} # 1) Validation Phase # We zip them to handle the parallel arrays easily for index, source, target in zip(indices, sources, targets): # Check if the substring in s matches the source # We use slicing to safely check bounds automatically if s[index : index + len(source)] == source: lookup[index] = (len(source), target) # 2) Construction Phase result = [] i = 0 while i < len(s): if i in lookup: # Found a replacement! source_len, target_str = lookup[i] result.append(target_str) i += source_len # Skip past the source characters else: # No replacement here, keep original character result.append(s[i]) i += 1 return ''.join(result)b7:T3538,

class Solution:
    def findReplaceString(self, s: str, indices: List[int], sources: List[str], targets: List[str]) -> str:
        # Map to store valid replacements: index -> (length_of_source, targets)
        lookup = {}

        # 1) Validation Phase
        # We zip them to handle the parallel arrays easily
        for index, source, target in zip(indices, sources, targets):
            # Check if the substring in s matches the source
            # We use slicing to safely check bounds automatically
            if s[index : index + len(source)] == source:
                lookup[index] = (len(source), target)
        # 2) Construction Phase
        result = []
        i = 0
        while i < len(s):
            if i in lookup:
                # Found a replacement!
                source_len, target_str = lookup[i]
                result.append(target_str)
                i += source_len # Skip past the source characters
            else:
                # No replacement here, keep original character
                result.append(s[i])
                i += 1
        return ''.join(result)

b8:T457,class Solution: def findReplaceString(self, s, indices, sources, targets): # 1. Sort all operations by index # We zip them to keep the data together, then sort based on the first item (index) sorted_ops = sorted(zip(indices, sources, targets)) result = [] last_pos = 0 for index, source, target in sorted_ops: # 2. Validation: Check if the source matches if s[index : index + len(source)] == source: # A. GAP FILL: Copy everything from the last position up to this index # This handles the "normal" characters in one go result.append(s[last_pos : index]) # B. REPLACE: Append the target result.append(target) # C. UPDATE POINTER: Move past the replaced part last_pos = index + len(source) # 3. Clean up: Append whatever is left after the final operation result.append(s[last_pos:]) return "".join(result)b9:T2a90,

class Solution:
    def findReplaceString(self, s, indices, sources, targets):
        # 1. Sort all operations by index
        # We zip them to keep the data together, then sort based on the first item (index)
        sorted_ops = sorted(zip(indices, sources, targets))
        
        result = []
        last_pos = 0
        
        for index, source, target in sorted_ops:
            # 2. Validation: Check if the source matches
            if s[index : index + len(source)] == source:
                
                # A. GAP FILL: Copy everything from the last position up to this index
                # This handles the "normal" characters in one go
                result.append(s[last_pos : index])
                
                # B. REPLACE: Append the target
                result.append(target)
                
                # C. UPDATE POINTER: Move past the replaced part
                last_pos = index + len(source)
                
        # 3. Clean up: Append whatever is left after the final operation
        result.append(s[last_pos:])
        
        return "".join(result)

ba:T2a28,

class Solution:
    def employeeFreeTime(self, schedule: '[[Interval]]') -> '[Interval]':
        # 1. Flatten the list of lists
        all_intervals = []
        for emp in schedule:
            for interval in emp:
                all_intervals.append(interval)
        
        # 2. Sort by start time (destroys the pre-sorted property, but it's fast)
        all_intervals.sort(key=lambda x: x.start)
        
        result = []
        prev_end = all_intervals[0].end
        
        # 3. Standard Sweep Line
        for i in range(1, len(all_intervals)):
            curr = all_intervals[i]
            
            if curr.start > prev_end:
                # Found a gap
                result.append(Interval(prev_end, curr.start))
                prev_end = curr.end
            else:
                # Overlap: extend our known busy time
                prev_end = max(prev_end, curr.end)
                
        return result

bb:T8a5,This approach is basically Merge K Sorted Lists (a classic hard pattern). The Concept: "Streaming" the Intervals Instead of dumping all intervals into a bucket and sorting them (which destroys the pre-sorted property given to us), we use a Min-Heap to keep track of the "next earliest available interval" among all employees. So like basically, rather than doing a "lets collect everything and mash it into one blob and sort it" we simply... go through the existing semi-sorted structure and continue to enforce the sortedness as we go along and collect everything. The Heap size will never exceed K (the number of employees). The Algorithm 1. Initialize: Add the first interval of every employee into a Min-Heap. 2. Heap Tuple: (start_time, employee_index, interval_index) 3. Track: Pop the smallest start time from the heap. This is our current interval. 4. Process: Compare with prev_end (just like the previous solution). 5. Refill: Once we process an interval from Employee X, grab Employee X's next interval and push it into the heap. Approach 1 (Flatten & Sort): "I don't trust the input. I will dump everything on the floor, mix it up, and re-organize it from scratch." Approach 2 (Heap): "I trust the input is locally sorted. I will just look at the front of each line and pick the winner." The Visual: The "Funnel" Think of the Min-Heap as a funnel. We pour K different sorted streams into the top, and the heap guarantees that what drops out of the bottom is always the single smallest element, one by one. We are "enforcing sortedness" by maintaining an invariant: The heap always holds the next available candidate from every employee. It never holds more than one interval per employee, so it stays small and fast. Bonus: The "Iterator" Pattern This is essentially the Iterator Pattern. We are creating a "Super Iterator" that wraps around K smaller iterators. Functions like Python's heapq.merge or Java's Guava Iterators.mergeSorted do exactly this under the hood. Knowing this problem well demonstrates that we understand how to build tools that process data lazily (only fetching the next item when asked), which is critical for massive distributed systems (like MapReduce).bc:T956,

This approach is basically Merge K Sorted Lists (a classic hard pattern).

The Concept: "Streaming" the Intervals

Instead of dumping all intervals into a bucket and sorting them (which destroys the pre-sorted property given to us), we use a Min-Heap to keep track of the "next earliest available interval" among all employees.

So like basically, rather than doing a "lets collect everything and mash it into one blob and sort it" we simply... go through the existing semi-sorted structure and continue to enforce the sortedness as we go along and collect everything.

The Heap size will never exceed K (the number of employees).

The Algorithm

Initialize: Add the first interval of every employee into a Min-Heap.
Heap Tuple: (start_time, employee_index, interval_index)
Track: Pop the smallest start time from the heap. This is our current interval.
Process: Compare with prev_end (just like the previous solution).
Refill: Once we process an interval from Employee X, grab Employee X's next interval and push it into the heap.

Approach 1 (Flatten & Sort): "I don't trust the input. I will dump everything on the floor, mix it up, and re-organize it from scratch."

Approach 2 (Heap): "I trust the input is locally sorted. I will just look at the front of each line and pick the winner."

The Visual: The "Funnel"

Think of the Min-Heap as a funnel. We pour K different sorted streams into the top, and the heap guarantees that what drops out of the bottom is always the single smallest element, one by one.

We are "enforcing sortedness" by maintaining an invariant: The heap always holds the next available candidate from every employee. It never holds more than one interval per employee, so it stays small and fast.

Bonus: The "Iterator" Pattern

This is essentially the Iterator Pattern.

We are creating a "Super Iterator" that wraps around K smaller iterators.

Functions like Python's heapq.merge or Java's Guava Iterators.mergeSorted do exactly this under the hood.

Knowing this problem well demonstrates that we understand how to build tools that process data lazily (only fetching the next item when asked), which is critical for massive distributed systems (like MapReduce).

bd:T792,class Solution: def employeeFreeTime(self, schedule: '[[Interval]]') -> '[Interval]': min_heap = [] # 1. Initialize Heap with the first interval of each employee # Heap Tuple: (start_time, employee_idx, interval_idx) for employee_idx, emp_schedule in enumerate(schedule): if emp_schedule: # We store the start time for sorting, and indices to find the next one start_time = emp_schedule[0].start heapq.heappush(min_heap, (start_time, employee_idx, 0)) result = [] # Initialize prev_end. We need to peek at the first real interval. if not min_heap: return [] # Peek at the first interval (employee_idx, 0) # We don't pop yet, just need to set the initial 'busy' boundary first_emp_idx = min_heap[0][1] prev_end = schedule[first_emp_idx][0].end # 2. Process the Heap while min_heap: # Pop the earliest starting interval start, emp_idx, interval_idx = heapq.heappop(min_heap) # Access the actual object using our indices curr_interval = schedule[emp_idx][interval_idx] # Check for Gap if start > prev_end: result.append(Interval(prev_end, start)) prev_end = curr_interval.end else: # Extend the busy block prev_end = max(prev_end, curr_interval.end) # 3. Refill the heap with THIS employee's next interval next_interval_idx = interval_idx + 1 if next_interval_idx < len(schedule[emp_idx]): next_interval = schedule[emp_idx][next_interval_idx] heapq.heappush(min_heap, (next_interval.start, emp_idx, next_interval_idx)) return resultbe:T447a,

class Solution:
    def employeeFreeTime(self, schedule: '[[Interval]]') -> '[Interval]':
        min_heap = []
        
        # 1. Initialize Heap with the first interval of each employee
        # Heap Tuple: (start_time, employee_idx, interval_idx)
        for employee_idx, emp_schedule in enumerate(schedule):
            if emp_schedule:
                # We store the start time for sorting, and indices to find the next one
                start_time = emp_schedule[0].start
                heapq.heappush(min_heap, (start_time, employee_idx, 0))
        
        result = []
        
        # Initialize prev_end. We need to peek at the first real interval.
        if not min_heap:
            return []
            
        # Peek at the first interval (employee_idx, 0)
        # We don't pop yet, just need to set the initial 'busy' boundary
        first_emp_idx = min_heap[0][1]
        prev_end = schedule[first_emp_idx][0].end
        
        # 2. Process the Heap
        while min_heap:
            # Pop the earliest starting interval
            start, emp_idx, interval_idx = heapq.heappop(min_heap)
            
            # Access the actual object using our indices
            curr_interval = schedule[emp_idx][interval_idx]
            
            # Check for Gap
            if start > prev_end:
                result.append(Interval(prev_end, start))
                prev_end = curr_interval.end
            else:
                # Extend the busy block
                prev_end = max(prev_end, curr_interval.end)
            
            # 3. Refill the heap with THIS employee's next interval
            next_interval_idx = interval_idx + 1
            if next_interval_idx < len(schedule[emp_idx]):
                next_interval = schedule[emp_idx][next_interval_idx]
                heapq.heappush(min_heap, (next_interval.start, emp_idx, next_interval_idx))
                
        return result

bf:Ta3d,**You rarely have to write the whole thing.** We can get away with writing 15 lines of code and "hand-waving" the rest. ### Pattern 1: The "Trie Skeleton" (The Muscle Memory) ```python class TrieNode: def __init__(self): self.children = {} # Map char -> Node self.is_word = False # Or store data here (like frequency) ``` ### Pattern 2: The "Crawler" (The Standard Loop) Whether you are inserting a word or searching for a word, the code is **identical**. You are just crawling down the tree. ```python node = root for char in word: if char not in node.children: # Create it (if inserting) OR Return False (if searching) node = node.children[char] # Now 'node' is at the end of the prefix ``` **Takeaway:** This loop is the answer to everything. * Want to insert? Use the loop. * Want to search? Use the loop. * Want autocomplete? Use the loop to get to the prefix node, *then* worry about the top 3. ### Pattern 3: The "Decoupled HashMap" (The Cheat Code) Here is how you make the code much, much simpler (and easier to write). Don't try to manage complex frequency counters *inside* the recursion. It's too messy. Instead, keep a separate **Global Hashmap** for counting. * **Trie:** Handles the structure (finding the path). * **HashMap:** Handles the math (counting frequencies). **How this saves you:** When the user types `#` (enter), you don't need to traverse the tree to update counts. You just do `global_map[sentence] += 1`. Done. 1. **Define the Class:** Write `class TrieNode`. 2. **Define the Dictionary:** Write `self.counts = {}`. 3. **Ask the "Lazy" Question:** > "To find the Top 3, is it okay if I just traverse the sub-tree (DFS) for now? We can optimize it with caching later if we have time." **95% of interviewers will say "Yes."** 4. **Write the DFS (if asked):** The DFS is just: "Visit my children. If I am a word, add me to the list." ### Summary **Problem:** Autocomplete / Typeahead **Data Structure:** Trie (Prefix Tree) + HashMap (for frequencies) **Key Concepts:** 1. **TrieNode:** `{ children: Map, is_end: Bool }` 2. **Input(char):** Traverse down one step. If valid, DFS from that point to gather results. 3. **Input(\#):** Update `HashMap[sentence] += 1` and `Trie.insert(sentence)`. 4. **Sorting:** `results.sort(key=lambda x: (-frequency, text))` **Trade-off (The "Bonus Points" discussion):** * **Current approach (DFS):** Slow Read (O(N)), Fast Write. Good for saving memory. * **Optimized approach (Cache at Node):** Fast Read (O(1)), Slow Write. Good for high-traffic search engines.c0:Td1c,

You rarely have to write the whole thing.

We can get away with writing 15 lines of code and "hand-waving" the rest.

Pattern 1: The "Trie Skeleton" (The Muscle Memory)

class TrieNode:
    def __init__(self):
        self.children = {}  # Map char -> Node
        self.is_word = False # Or store data here (like frequency)

Pattern 2: The "Crawler" (The Standard Loop)

Whether you are inserting a word or searching for a word, the code is identical. You are just crawling down the tree.

node = root
for char in word:
    if char not in node.children:
        # Create it (if inserting) OR Return False (if searching)
    node = node.children[char]
# Now 'node' is at the end of the prefix

Takeaway: This loop is the answer to everything.

Want to insert? Use the loop.
Want to search? Use the loop.
Want autocomplete? Use the loop to get to the prefix node, then worry about the top 3.

Pattern 3: The "Decoupled HashMap" (The Cheat Code)

Here is how you make the code much, much simpler (and easier to write).

Don't try to manage complex frequency counters inside the recursion. It's too messy.
Instead, keep a separate Global Hashmap for counting.

Trie: Handles the structure (finding the path).
HashMap: Handles the math (counting frequencies).

How this saves you:
When the user types # (enter), you don't need to traverse the tree to update counts. You just do global_map[sentence] += 1. Done.

Define the Class: Write class TrieNode.
Define the Dictionary: Write self.counts = {}.
Ask the "Lazy" Question:

"To find the Top 3, is it okay if I just traverse the sub-tree (DFS) for now? We can optimize it with caching later if we have time."

95% of interviewers will say "Yes."
Write the DFS (if asked):
The DFS is just: "Visit my children. If I am a word, add me to the list."

Summary

Problem: Autocomplete / Typeahead
Data Structure: Trie (Prefix Tree) + HashMap (for frequencies)

Key Concepts:

TrieNode: { children: Map, is_end: Bool }
Input(char): Traverse down one step. If valid, DFS from that point to gather results.
Input(#): Update HashMap[sentence] += 1 and Trie.insert(sentence).
Sorting: results.sort(key=lambda x: (-frequency, text))

Trade-off (The "Bonus Points" discussion):

Current approach (DFS): Slow Read (O(N)), Fast Write. Good for saving memory.
Optimized approach (Cache at Node): Fast Read (O(1)), Slow Write. Good for high-traffic search engines.

c1:T9b2,class TrieNode: def __init__(self): self.children = {} self.is_end = False # We store the full sentence at the leaf to make retrieval easier self.sentence = None # We can store frequency at the leaf, # but a global hashmap is often cleaner for updates. class AutocompleteSystem: def __init__(self, sentences: list[str], times: list[int]): self.root = TrieNode() self.current_sentence = "" self.keyword_counts = {} # Global map for frequency: sentence -> count # Initialize the system with historical data for i in range(len(sentences)): self.keyword_counts[sentences[i]] = times[i] self.add_to_trie(sentences[i]) def add_to_trie(self, sentence): node = self.root for char in sentence: if char not in node.children: node.children[char] = TrieNode() node = node.children[char] node.is_end = True node.sentence = sentence def input(self, c: str) -> list[str]: # CASE 1: End of sentence if c == '#': # Update frequency self.keyword_counts[self.current_sentence] = self.keyword_counts.get(self.current_sentence, 0) + 1 # Ensure it's in the trie self.add_to_trie(self.current_sentence) # Reset buffer and return empty self.current_sentence = "" return [] # CASE 2: Typing a character self.current_sentence += c # Traverse to the current prefix node node = self.root for char in self.current_sentence: if char not in node.children: return [] # Prefix doesn't exist node = node.children[char] # Perform DFS to find all sentences starting with this prefix results = [] self.dfs(node, results) # Sort results: # 1. Frequency (descending) -> -x[1] # 2. ASCII (ascending) -> x[0] results.sort(key=lambda x: (-x[1], x[0])) # Return top 3 sentences (extracting just the string) return [item[0] for item in results[:3]] def dfs(self, node, results): if node.is_end: count = self.keyword_counts[node.sentence] results.append((node.sentence, count)) for child in node.children.values(): self.dfs(child, results)c2:T8180,

class TrieNode:
    def __init__(self):
        self.children = {}
        self.is_end = False
        # We store the full sentence at the leaf to make retrieval easier
        self.sentence = None 
        # We can store frequency at the leaf, 
        # but a global hashmap is often cleaner for updates.

class AutocompleteSystem:

    def __init__(self, sentences: list[str], times: list[int]):
        self.root = TrieNode()
        self.current_sentence = ""
        self.keyword_counts = {}  # Global map for frequency: sentence -> count
        
        # Initialize the system with historical data
        for i in range(len(sentences)):
            self.keyword_counts[sentences[i]] = times[i]
            self.add_to_trie(sentences[i])

    def add_to_trie(self, sentence):
        node = self.root
        for char in sentence:
            if char not in node.children:
                node.children[char] = TrieNode()
            node = node.children[char]
        node.is_end = True
        node.sentence = sentence

    def input(self, c: str) -> list[str]:
        # CASE 1: End of sentence
        if c == '#':
            # Update frequency
            self.keyword_counts[self.current_sentence] = self.keyword_counts.get(self.current_sentence, 0) + 1
            # Ensure it's in the trie
            self.add_to_trie(self.current_sentence)
            # Reset buffer and return empty
            self.current_sentence = ""
            return []
        
        # CASE 2: Typing a character
        self.current_sentence += c
        
        # Traverse to the current prefix node
        node = self.root
        for char in self.current_sentence:
            if char not in node.children:
                return [] # Prefix doesn't exist
            node = node.children[char]
            
        # Perform DFS to find all sentences starting with this prefix
        results = []
        self.dfs(node, results)
        
        # Sort results: 
        # 1. Frequency (descending) -> -x[1]
        # 2. ASCII (ascending) -> x[0]
        results.sort(key=lambda x: (-x[1], x[0]))
        
        # Return top 3 sentences (extracting just the string)
        return [item[0] for item in results[:3]]

    def dfs(self, node, results):
        if node.is_end:
            count = self.keyword_counts[node.sentence]
            results.append((node.sentence, count))
            
        for child in node.children.values():
            self.dfs(child, results)

c3:T6e0,class TrieNode: def __init__(self): self.children = {} self.is_file = False # Use list for O(1) appends, join only on read self.content = [] class FileSystem: def __init__(self): self.root = TrieNode() def _traverse(self, path: str, create: bool = False): """ Unified helper for navigation. Returns the node if found/created, else None. """ node = self.root # Split by '/' and ignore empty strings (handles "//" or trailing "/") parts = [p for p in path.split('/') if p] for part in parts: if part not in node.children: if not create: return None node.children[part] = TrieNode() node = node.children[part] return node def ls(self, path: str) -> List[str]: node = self._traverse(path) if not node: return [] if node.is_file: # We guaranteed root can't be a file, so split works safely return [path.split('/')[-1]] return sorted(node.children.keys()) def mkdir(self, path: str) -> None: self._traverse(path, create=True) def addContentToFile(self, filePath: str, content: str) -> None: # Guard clause: Prevent corrupting the root directory if filePath == "/": return node = self._traverse(filePath, create=True) node.is_file = True node.content.append(content) def readContentFromFile(self, filePath: str) -> str: node = self._traverse(filePath) if not node or not node.is_file: return "" return "".join(node.content)c4:T62b7,

class TrieNode:
    def __init__(self):
        self.children = {}
        self.is_file = False
        # Use list for O(1) appends, join only on read
        self.content = []

class FileSystem:

    def __init__(self):
        self.root = TrieNode()

    def _traverse(self, path: str, create: bool = False):
        """
        Unified helper for navigation. 
        Returns the node if found/created, else None.
        """
        node = self.root
        # Split by '/' and ignore empty strings (handles "//" or trailing "/")
        parts = [p for p in path.split('/') if p]

        for part in parts:
            if part not in node.children:
                if not create:
                    return None
                node.children[part] = TrieNode()
            node = node.children[part]
        return node

    def ls(self, path: str) -> List[str]:
        node = self._traverse(path)
        
        if not node:
            return []
        
        if node.is_file:
            # We guaranteed root can't be a file, so split works safely
            return [path.split('/')[-1]]

        return sorted(node.children.keys())

    def mkdir(self, path: str) -> None:
        self._traverse(path, create=True)

    def addContentToFile(self, filePath: str, content: str) -> None:
        # Guard clause: Prevent corrupting the root directory
        if filePath == "/":
            return 
            
        node = self._traverse(filePath, create=True)
        node.is_file = True
        node.content.append(content)

    def readContentFromFile(self, filePath: str) -> str:
        node = self._traverse(filePath)
        
        if not node or not node.is_file:
            return ""
            
        return "".join(node.content)

c5:T18bb,

import heapq

def top_k_elements(nums, k):
    # 1. Count frequencies (if needed)
    count = {} # create hashmap of frequencies
    
    # 2. The Heap
    heap = [] # Min-heap stores (frequency, value)
    
    for num, freq in count.items():
        # Push into heap
        heapq.heappush(heap, (freq, num))
        
        # 3. Maintain size K
        if len(heap) > k:
            heapq.heappop(heap) # Removes the SMALLEST frequency
            
    return [val for freq, val in heap]

c6:T20cc,

def sliding_window(s):
    left = 0
    window_state = {} # Or set(), or integer sum
    best_len = 0
    
    # 1. Expand (Right pointer moves)
    for right in range(len(s)):
        char = s[right]
        window_state[char] = window_state.get(char, 0) + 1
        
        # 2. Shrink (While invalid)
        # Example: If we have duplicates
        while window_state[char] > 1:
            left_char = s[left]
            window_state[left_char] -= 1
            left += 1
        
        # 3. Update Result (After fixing the window)
        best_len = max(best_len, right - left + 1)
        
    return best_len

c7:T224d,

def min_capacity(weights, days):
    # 1. Define Search Space
    # The answer is somewhere between max(weights) and sum(weights)
    left, right = max(weights), sum(weights)
    
    # 2. Feasibility Function
    def can_ship(capacity):
        # Logic to check if this specific capacity works
        ships_needed = 1
        current_load = 0
        for w in weights:
            if current_load + w > capacity:
                ships_needed += 1
                current_load = 0
            current_load += w
        return ships_needed <= days

    # 3. Binary Search Template
    while left < right:
        mid = (left + right) // 2
        if can_ship(mid):
            # If it works, try to find a smaller/better answer
            right = mid
        else:
            # If it fails, we strictly need more capacity
            left = mid + 1
            
    return left

c8:T38de,

def solve_islands(grid):
    rows, cols = len(grid), len(grid[0])
    visited = set() # Or modify grid directly to '#' to save space

    def dfs(r, c):
        # 1. Base Case: Out of bounds or already visited/invalid
        if (r < 0 or r >= rows or c < 0 or c >= cols or
            (r, c) in visited or grid[r][c] == "0"):
            return 0
        
        # 2. Mark visited
        visited.add((r, c))
        
        # 3. Recurse (Summing results if needed, e.g., area)
        size = 1
        size += dfs(r + 1, c)
        size += dfs(r - 1, c)
        size += dfs(r, c + 1)
        size += dfs(r, c - 1)
        return size

    # 4. Main loop to find disconnected components
    max_area = 0
    for r in range(rows):
        for c in range(cols):
            if grid[r][c] == "1" and (r, c) not in visited:
                max_area = max(max_area, dfs(r, c))
    return max_area

c9:T4a1,from collections import deque def bfs_matrix(grid): rows, cols = len(grid), len(grid[0]) queue = deque() visited = set() # 1. Initialize (Add all starting points) # Example: Find all '2's (rotten oranges) for r in range(rows): for c in range(cols): if grid[r][c] == 2: queue.append((r, c)) visited.add((r, c)) steps = 0 directions = [(0, 1), (0, -1), (1, 0), (-1, 0)] # Right, Left, Down, Up # 2. The Loop while queue: # Range(len) is crucial to distinguish "layers" (distance) for _ in range(len(queue)): r, c = queue.popleft() # 3. Check neighbors for dr, dc in directions: nr, nc = r + dr, c + dc # 4. Validate Bounds & Visited if (0 <= nr < rows and 0 <= nc < cols and (nr, nc) not in visited and grid[nr][nc] == 1): # Example condition: Fresh orange visited.add((nr, nc)) queue.append((nr, nc)) steps += 1 return stepsca:T4291,

from collections import deque

def bfs_matrix(grid):
    rows, cols = len(grid), len(grid[0])
    queue = deque()
    visited = set()
    
    # 1. Initialize (Add all starting points)
    # Example: Find all '2's (rotten oranges)
    for r in range(rows):
        for c in range(cols):
            if grid[r][c] == 2:
                queue.append((r, c))
                visited.add((r, c))
    
    steps = 0
    directions = [(0, 1), (0, -1), (1, 0), (-1, 0)] # Right, Left, Down, Up
    
    # 2. The Loop
    while queue:
        # Range(len) is crucial to distinguish "layers" (distance)
        for _ in range(len(queue)):
            r, c = queue.popleft()
            
            # 3. Check neighbors
            for dr, dc in directions:
                nr, nc = r + dr, c + dc
                
                # 4. Validate Bounds & Visited
                if (0 <= nr < rows and 0 <= nc < cols and 
                    (nr, nc) not in visited and 
                    grid[nr][nc] == 1): # Example condition: Fresh orange
                    
                    visited.add((nr, nc))
                    queue.append((nr, nc))
        steps += 1
        
    return steps

cb:T40e,

The Setup (Constructor)
You need an array parent. Initially, everyone is their own parent.
Mnemonic: "I am my own boss."

The find function (Path Compression)
This is the most important part.
Logic: If I am not my own boss, go find my boss's boss, and save that result so I don't have to ask again next time.
Template: Recursive is easiest to remember.

Version A: The "Lazy" Union (No Rank)
Use this when you are short on time or the constraints are small (N<10,000). It just blindly sets one parent to the other.

Version B: The "Pro" Union (With Rank)
Use this for hard graph problems (N>100,000) or to impress an interviewer.
Logic: Find the roots. If they are different, check who is bigger (higher rank). The small one points to the big one. If they are equal, pick one to win and increase its rank.

https://aistudio.google.com/prompts/1Yj7xbJdZW_gmgxOCkb2ozD_4ER4HlWXX

cc:T2e83,

class DSU:
    def __init__(self, n):
        self.parent = list(range(n))
        # self.rank = [1] * n  <-- Optional: Uncomment for efficiency

    def find(self, i):
        if self.parent[i] != i:
            self.parent[i] = self.find(self.parent[i])
        return self.parent[i]

    def union(self, i, j):
        root_i = self.find(i)
        root_j = self.find(j)
        
        if root_i != root_j:
            # Simple version (usually sufficient):
            self.parent[root_i] = root_j
            
            # Optimized version (replace line above with this):
            # if self.rank[root_i] > self.rank[root_j]:
            #     self.parent[root_j] = root_i
            # elif self.rank[root_j] > self.rank[root_i]:
            #     self.parent[root_i] = root_j
            # else:
            #     self.parent[root_j] = root_i
            #     self.rank[root_i] += 1
            return True
        return False

cd:T7c9,import heapq from collections import defaultdict from typing import List class Solution: def maxProbability(self, n: int, edges: List[List[int]], succProb: List[float], start_node: int, end_node: int) -> float: # 1. Build the Graph (Adjacency List) # Since the graph is undirected, we add edges for both directions. graph = defaultdict(list) for i, (u, v) in enumerate(edges): prob = succProb[i] graph[u].append((v, prob)) graph[v].append((u, prob)) # 2. Initialize Dijkstra's Variables # Max-Heap stores: (-probability, current_node) # We start with probability 1.0 at the start_node (stored as -1.0) pq = [(-1.0, start_node)] # Array to store the max probability to reach node 'i' max_prob = [0.0] * n max_prob[start_node] = 1.0 # 3. Process the Heap while pq: cur_p, cur_node = heapq.heappop(pq) cur_p = -cur_p # Convert back to positive # If we reached the target, this is guaranteed to be the max probability # due to the greedy nature of Dijkstra's algorithm. if cur_node == end_node: return cur_p # Optimization: If the current probability is smaller than what we # have already found for this node, skip processing. if cur_p < max_prob[cur_node]: continue # Explore neighbors for neighbor, edge_prob in graph[cur_node]: new_prob = cur_p * edge_prob # If we found a path with a higher probability, update and push to heap if new_prob > max_prob[neighbor]: max_prob[neighbor] = new_prob heapq.heappush(pq, (-new_prob, neighbor)) # If the queue empties and we never reached end_node return 0.0ce:T4ebc,

import heapq
from collections import defaultdict
from typing import List

class Solution:
    def maxProbability(self, n: int, edges: List[List[int]], succProb: List[float], start_node: int, end_node: int) -> float:
        # 1. Build the Graph (Adjacency List)
        # Since the graph is undirected, we add edges for both directions.
        graph = defaultdict(list)
        for i, (u, v) in enumerate(edges):
            prob = succProb[i]
            graph[u].append((v, prob))
            graph[v].append((u, prob))
        
        # 2. Initialize Dijkstra's Variables
        # Max-Heap stores: (-probability, current_node)
        # We start with probability 1.0 at the start_node (stored as -1.0)
        pq = [(-1.0, start_node)]
        
        # Array to store the max probability to reach node 'i'
        max_prob = [0.0] * n
        max_prob[start_node] = 1.0
        
        # 3. Process the Heap
        while pq:
            cur_p, cur_node = heapq.heappop(pq)
            cur_p = -cur_p  # Convert back to positive
            
            # If we reached the target, this is guaranteed to be the max probability
            # due to the greedy nature of Dijkstra's algorithm.
            if cur_node == end_node:
                return cur_p
            
            # Optimization: If the current probability is smaller than what we 
            # have already found for this node, skip processing.
            if cur_p < max_prob[cur_node]:
                continue
            
            # Explore neighbors
            for neighbor, edge_prob in graph[cur_node]:
                new_prob = cur_p * edge_prob
                
                # If we found a path with a higher probability, update and push to heap
                if new_prob > max_prob[neighbor]:
                    max_prob[neighbor] = new_prob
                    heapq.heappush(pq, (-new_prob, neighbor))
        
        # If the queue empties and we never reached end_node
        return 0.0

cf:T4f8,

Why this works (The Logic)

Imagine the lists look like this:

List A: [non-shared A] + [shared]
List B: [non-shared B] + [shared]

If you conceptually concatenate the two lists, you create two paths of equal length:

Path A + B: [non-shared A] $\to$ [shared] $\to$ [non-shared B] $\to$ [shared]
Path B + A: [non-shared B] $\to$ [shared] $\to$ [non-shared A] $\to$ [shared]

By switching pointers to the head of the other list once they hit the end, both pointers will travel exactly len(A) + len(B) nodes. Because the total distance is now the same, they align perfectly at the intersection node.

Edge Case: No Intersection

If there is no intersection, both pointers will traverse List A and then List B (and vice versa) and finally become None at the exact same time. The loop while pA != pB terminates because None == None, and None is returned.

d0:T1b2b,

class Solution:
    def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> Optional[ListNode]:
        if not headA or not headB:
            return None
        
        pA, pB = headA, headB
        
        while pA != pB:
            # If pA reaches the end of list A, redirect it to the head of list B
            # Otherwise, move to the next node
            pA = headB if pA is None else pA.next
            
            # If pB reaches the end of list B, redirect it to the head of list A
            # Otherwise, move to the next node
            pB = headA if pB is None else pB.next
            
        # Either they meet at the intersection node, or they meet at None (no intersection)
        return pA

d1:T83b,Bottom-Up Dynamic Programming. Sorting: By sorting words by length, we satisfy the topological order requirement. We never need to look "forward" in the list, only backward at shorter words we've already processed. Generation vs. Comparison: instead of comparing every word against every other word (O(N^2)), we take a word and generate all possible predecessors by removing one character. Since the maximum length (L) is only 16, generating predecessors is extremely fast. In Longest String Chain, you are given a simple list of words (e.g., ["a", "b", "ba", "bca", "bda", "bdca"]). It looks like an array problem, not a graph problem. You have to realize that words are nodes and a "predecessor" relationship (deleting one character) defines the edges. "How to extend this to a big-data situation?" This is a standard Google transition. They want to know if you understand that you cannot store a massive adjacency list in the RAM of a single machine. The "Correct" Practical Answer: Storage: You cannot store the graph in memory. You store the adjacency list in a distributed Key-Value store (like BigTable or Cassandra). Key = Node ID, Value = List of Neighbors. Processing: If you need a specific path (BFS): You use a distributed queue (like Kafka) to manage the "frontier" of nodes to visit. If you need to process the whole graph (like Longest Chain): You use MapReduce or Spark. Map Step: Emit key-value pairs for edges. Reduce Step: Aggregate connections and calculate path lengths iteratively. The practical answer leverages MapReduce logic: 1. Bucket by Length: You cannot find a predecessor for a 5-letter word anywhere except in the 4-letter bucket. 2. Distributed Processing: You can process all 4-letter words on one server and all 5-letter words on another. 3. The MapReduce Job: - Map: For every word of length $K$ (e.g., "apple"), generate all possible predecessors of length $K-1$ ("pple", "aple"...) as keys, with the value being the original word. - Join/Reduce: Join these generated keys with the actual list of $K-1$ words to verify which connections actually exist.d2:T8e2,

Bottom-Up Dynamic Programming.

Sorting: By sorting words by length, we satisfy the topological order requirement. We never need to look "forward" in the list, only backward at shorter words we've already processed.

Generation vs. Comparison: instead of comparing every word against every other word (O(N^2)), we take a word and generate all possible predecessors by removing one character.

Since the maximum length (L) is only 16, generating predecessors is extremely fast.

In Longest String Chain, you are given a simple list of words (e.g., ["a", "b", "ba", "bca", "bda", "bdca"]). It looks like an array problem, not a graph problem. You have to realize that words are nodes and a "predecessor" relationship (deleting one character) defines the edges.

"How to extend this to a big-data situation?"

This is a standard Google transition. They want to know if you understand that you cannot store a massive adjacency list in the RAM of a single machine.

The "Correct" Practical Answer:

Storage: You cannot store the graph in memory. You store the adjacency list in a distributed Key-Value store (like BigTable or Cassandra). Key = Node ID, Value = List of Neighbors.

Processing:

If you need a specific path (BFS): You use a distributed queue (like Kafka) to manage the "frontier" of nodes to visit.

If you need to process the whole graph (like Longest Chain): You use MapReduce or Spark.

Map Step: Emit key-value pairs for edges.

Reduce Step: Aggregate connections and calculate path lengths iteratively.

The practical answer leverages MapReduce logic:

Bucket by Length: You cannot find a predecessor for a 5-letter word anywhere except in the 4-letter bucket.
Distributed Processing: You can process all 4-letter words on one server and all 5-letter words on another.
The MapReduce Job:

Map: For every word of length $K$ (e.g., "apple"), generate all possible predecessors of length $K-1$ ("pple", "aple"...) as keys, with the value being the original word.
Join/Reduce: Join these generated keys with the actual list of $K-1$ words to verify which connections actually exist.

d3:T46f,class Solution: def longestStrChain(self, words: List[str]) -> int: # 1. Sort by length so we process shorter words first. # This ensures that when we process a word, its potential # predecessors have already been computed. words.sort(key=len) # 2. Dictionary to store the max chain length ending at each word. dp = {} max_chain = 1 for word in words: # Initialize current word's chain length to 1 (the word itself) dp[word] = 1 # 3. Try removing one character at every position to generate # potential predecessors. for i in range(len(word)): prev_word = word[:i] + word[i+1:] # 4. If the generated predecessor exists in our DP map, # update the chain length. if prev_word in dp: dp[word] = max(dp[word], dp[prev_word] + 1) # Track the global maximum max_chain = max(max_chain, dp[word]) return max_chaind4:T2e9a,

class Solution:
    def longestStrChain(self, words: List[str]) -> int:
        # 1. Sort by length so we process shorter words first.
        #    This ensures that when we process a word, its potential
        #    predecessors have already been computed.
        words.sort(key=len)
        
        # 2. Dictionary to store the max chain length ending at each word.
        dp = {}
        max_chain = 1
        
        for word in words:
            # Initialize current word's chain length to 1 (the word itself)
            dp[word] = 1
            
            # 3. Try removing one character at every position to generate
            #    potential predecessors.
            for i in range(len(word)):
                prev_word = word[:i] + word[i+1:]
                
                # 4. If the generated predecessor exists in our DP map,
                #    update the chain length.
                if prev_word in dp:
                    dp[word] = max(dp[word], dp[prev_word] + 1)
            
            # Track the global maximum
            max_chain = max(max_chain, dp[word])
            
        return max_chain

d5:T774,class Solution: def minTransfers(self, transactions: List[List[int]]) -> int: # 1. Calculate net balance for each person person_balance = defaultdict(int) for u, v, amount in transactions: person_balance[u] -= amount person_balance[v] += amount # 2. Filter out people who are already settled (0 balance) # We only care about non-zero balances. debts = [b for b in person_balance.values() if b != 0] return self.dfs(0, debts) def dfs(self, start_index: int, debts: List[int]) -> int: # Skip people who have already been settled (balance is 0) while start_index < len(debts) and debts[start_index] == 0: start_index += 1 # Base case: If we've gone through everyone, 0 transactions needed if start_index == len(debts): return 0 min_transactions = float('inf') # Try to settle the current person's debt (debts[start_index]) # by combining it with a subsequent person (debts[i]) for i in range(start_index + 1, len(debts)): # Optimization: Only attempt to settle if they have opposite signs. # (One owes money, the other is owed money). if debts[start_index] * debts[i] < 0: # "Transfer" balance from start_index to i. # We effectively zero out debts[start_index] and add it to debts[i]. debts[i] += debts[start_index] # 1 transaction + result of the recursive call min_transactions = min(min_transactions, 1 + self.dfs(start_index + 1, debts)) # Backtrack: Undo the transfer for the next iteration debts[i] -= debts[start_index] return min_transactionsd6:T4f6c,

class Solution:
    def minTransfers(self, transactions: List[List[int]]) -> int:
        # 1. Calculate net balance for each person
        person_balance = defaultdict(int)
        for u, v, amount in transactions:
            person_balance[u] -= amount
            person_balance[v] += amount
        
        # 2. Filter out people who are already settled (0 balance)
        # We only care about non-zero balances.
        debts = [b for b in person_balance.values() if b != 0]
        
        return self.dfs(0, debts)

    def dfs(self, start_index: int, debts: List[int]) -> int:
        # Skip people who have already been settled (balance is 0)
        while start_index < len(debts) and debts[start_index] == 0:
            start_index += 1
        
        # Base case: If we've gone through everyone, 0 transactions needed
        if start_index == len(debts):
            return 0
        
        min_transactions = float('inf')
        
        # Try to settle the current person's debt (debts[start_index])
        # by combining it with a subsequent person (debts[i])
        for i in range(start_index + 1, len(debts)):
            
            # Optimization: Only attempt to settle if they have opposite signs.
            # (One owes money, the other is owed money).
            if debts[start_index] * debts[i] < 0:
                
                # "Transfer" balance from start_index to i. 
                # We effectively zero out debts[start_index] and add it to debts[i].
                debts[i] += debts[start_index]
                
                # 1 transaction + result of the recursive call
                min_transactions = min(min_transactions, 1 + self.dfs(start_index + 1, debts))
                
                # Backtrack: Undo the transfer for the next iteration
                debts[i] -= debts[start_index]
                
        return min_transactions

d7:T3102,

class Solution:
    def findMaxForm(self, strs: list[str], m: int, n: int) -> int:
        # dp[i][j] represents the max subset size with i zeros and j ones
        # Dimensions are (m+1) x (n+1) initialized to 0
        dp = [[0] * (n + 1) for _ in range(m + 1)]
        
        for s in strs:
            # Count costs for the current string
            zeros = s.count('0')
            ones = len(s) - zeros
            
            # Iterate backwards through the DP matrix to ensure 
            # we only use the current string once per state.
            # We stop when i < zeros or j < ones because the string
            # cannot fit in the remaining capacity.
            for i in range(m, zeros - 1, -1):
                for j in range(n, ones - 1, -1):
                    dp[i][j] = max(dp[i][j], 1 + dp[i - zeros][j - ones])
                    
        return dp[m][n]

d8:T1c47,

class Solution:
    def getImportance(self, employees: List['Employee'], id: int) -> int:
        # 1. Create a hash map for O(1) lookup: {id -> Employee Object}
        emap = {e.id: e for e in employees}
        
        # 2. Define a helper function for Depth First Search
        def dfs(eid):
            employee = emap[eid]
            # Return current importance + sum of subordinates' importance
            return employee.importance + sum(dfs(sub_id) for sub_id in employee.subordinates)
        
        # 3. Kick off the recursion
        return dfs(id)

d9:T428,class Solution: def pathSum(self, root: Optional[TreeNode], targetSum: int) -> List[List[int]]: result = [] def dfs(node, remaining_sum, current_path): if not node: return # 1. Add current node to path and update sum current_path.append(node.val) # 2. Check if it's a leaf node and sum matches if not node.left and not node.right and remaining_sum == node.val: # Important: Append a copy ([:]) of the list, # otherwise backtracks will affect the result. result.append(current_path[:]) # 3. Traverse children else: dfs(node.left, remaining_sum - node.val, current_path) dfs(node.right, remaining_sum - node.val, current_path) # 4. Backtrack: Remove current node to prepare for the next branch current_path.pop() dfs(root, targetSum, []) return resultda:T2d63,

class Solution:
    def pathSum(self, root: Optional[TreeNode], targetSum: int) -> List[List[int]]:
        result = []
        
        def dfs(node, remaining_sum, current_path):
            if not node:
                return
            
            # 1. Add current node to path and update sum
            current_path.append(node.val)
            
            # 2. Check if it's a leaf node and sum matches
            if not node.left and not node.right and remaining_sum == node.val:
                # Important: Append a copy ([:]) of the list, 
                # otherwise backtracks will affect the result.
                result.append(current_path[:])
            
            # 3. Traverse children
            else:
                dfs(node.left, remaining_sum - node.val, current_path)
                dfs(node.right, remaining_sum - node.val, current_path)
            
            # 4. Backtrack: Remove current node to prepare for the next branch
            current_path.pop()
            
        dfs(root, targetSum, [])
        return result

db:T2367,

class Solution:
    def minimumTotal(self, triangle: List[List[int]]) -> int:
        # Initialize dp array with the values of the bottom row
        dp = triangle[-1][:]

        # Iterate from the second-to-last row up to the top row
        for r in range(len(triangle) - 2, -1, -1):
            for c in range(len(triangle[r])):
                # The minimum path sum at this specific node is:
                # its own value + the minimum of the two nodes below it
                dp[c] = triangle[r][c] + min(dp[c], dp[c + 1])

        # The top element now contains the minimum path sum for the whole triangle
        return dp[0]

dc:T1d77,

class Solution:
    def minimumTotal(self, triangle: List[List[int]]) -> int:
        for r in range(len(triangle) - 2, -1, -1):
            for c in range(len(triangle[r])):
                triangle[r][c] += min(triangle[r+1][c], triangle[r+1][c+1])
        return triangle[0][0]

dd:T2887,

class Solution:
    def PredictTheWinner(self, nums: List[int]) -> bool:
        n = len(nums)
        # dp[i] will eventually store the max diff for subarray ending at index j
        # initialized with the base case (subarrays of length 1)
        dp = nums[:] 
        
        # Iterate over the length of the subarray
        # diff is the distance between left pointer (i) and right pointer (j)
        for diff in range(1, n):
            for i in range(n - diff):
                j = i + diff
                # Update dp[i] for the new range [i, j]
                # dp[i+1] represents the result of range [i+1, j] (after picking left)
                # dp[i] (old value) represents the result of range [i, j-1] (after picking right)
                pick_left = nums[i] - dp[i + 1]
                pick_right = nums[j] - dp[i]
                
                dp[i] = max(pick_left, pick_right)
        
        return dp[0] >= 0

de:T593,class Bank: def __init__(self, balance: List[int]): # Store the balance list. # We keep the length to avoid calling len() repeatedly in checks. self.balance = balance self.n = len(balance) def transfer(self, account1: int, account2: int, money: int) -> bool: # Check if both accounts exist (1 <= account <= n) if not (1 <= account1 <= self.n and 1 <= account2 <= self.n): return False # Check if sender has enough money # Note: account numbers are 1-indexed, list is 0-indexed if self.balance[account1 - 1] < money: return False # Perform transfer self.balance[account1 - 1] -= money self.balance[account2 - 1] += money return True def deposit(self, account: int, money: int) -> bool: # Check if account exists if not (1 <= account <= self.n): return False # Perform deposit self.balance[account - 1] += money return True def withdraw(self, account: int, money: int) -> bool: # Check if account exists if not (1 <= account <= self.n): return False # Check if account has enough money if self.balance[account - 1] < money: return False # Perform withdrawal self.balance[account - 1] -= money return Truedf:T52e5,

class Bank:
    def __init__(self, balance: List[int]):
        # Store the balance list. 
        # We keep the length to avoid calling len() repeatedly in checks.
        self.balance = balance
        self.n = len(balance)

    def transfer(self, account1: int, account2: int, money: int) -> bool:
        # Check if both accounts exist (1 <= account <= n)
        if not (1 <= account1 <= self.n and 1 <= account2 <= self.n):
            return False
        
        # Check if sender has enough money
        # Note: account numbers are 1-indexed, list is 0-indexed
        if self.balance[account1 - 1] < money:
            return False
        
        # Perform transfer
        self.balance[account1 - 1] -= money
        self.balance[account2 - 1] += money
        return True

    def deposit(self, account: int, money: int) -> bool:
        # Check if account exists
        if not (1 <= account <= self.n):
            return False
        
        # Perform deposit
        self.balance[account - 1] += money
        return True

    def withdraw(self, account: int, money: int) -> bool:
        # Check if account exists
        if not (1 <= account <= self.n):
            return False
        
        # Check if account has enough money
        if self.balance[account - 1] < money:
            return False
        
        # Perform withdrawal
        self.balance[account - 1] -= money
        return True

e0:T54f,class Solution: def earliestAcq(self, logs: List[List[int]], n: int) -> int: # 1. Sort logs by timestamp (index 0) to ensure chronological order logs.sort(key=lambda x: x[0]) # Initialize Union-Find structures # parent[i] points to the parent of node i parent = list(range(n)) # group_count tracks the number of disjoint sets group_count = n # Helper: Find the representative (root) of group x with Path Compression def find(x): if parent[x] != x: parent[x] = find(parent[x]) return parent[x] # Helper: Unite two groups. Returns True if a merge happened. def union(x, y): root_x = find(x) root_y = find(y) if root_x != root_y: # Merge the sets (arbitrarily attach y to x) parent[root_y] = root_x return True return False # 2. Iterate through sorted logs for timestamp, x, y in logs: if union(x, y): group_count -= 1 # 3. Check if everyone is in one group if group_count == 1: return timestamp # If we process all logs and still have > 1 group return -1e1:T3b3e,

class Solution:
    def earliestAcq(self, logs: List[List[int]], n: int) -> int:
        # 1. Sort logs by timestamp (index 0) to ensure chronological order
        logs.sort(key=lambda x: x[0])
        
        # Initialize Union-Find structures
        # parent[i] points to the parent of node i
        parent = list(range(n))
        # group_count tracks the number of disjoint sets
        group_count = n
        
        # Helper: Find the representative (root) of group x with Path Compression
        def find(x):
            if parent[x] != x:
                parent[x] = find(parent[x])
            return parent[x]
        
        # Helper: Unite two groups. Returns True if a merge happened.
        def union(x, y):
            root_x = find(x)
            root_y = find(y)
            
            if root_x != root_y:
                # Merge the sets (arbitrarily attach y to x)
                parent[root_y] = root_x
                return True
            return False

        # 2. Iterate through sorted logs
        for timestamp, x, y in logs:
            if union(x, y):
                group_count -= 1
                
                # 3. Check if everyone is in one group
                if group_count == 1:
                    return timestamp
        
        # If we process all logs and still have > 1 group
        return -1

e2:T233a,

class Solution:
    def maxScore(self, cardPoints: List[int], k: int) -> int:
        # 1. Start by taking all k cards from the left side.
        left_sum = sum(cardPoints[:k])
        right_sum = 0
        max_points = left_sum
        
        n = len(cardPoints)
        
        # 2. Slide the window: give one up from the left, take one from the right.
        for i in range(k):
            # Remove the last card we took from the left
            left_sum -= cardPoints[k - 1 - i]
            # Add a card from the end of the array (right side)
            right_sum += cardPoints[n - 1 - i]
            
            # Update maximum found so far
            max_points = max(max_points, left_sum + right_sum)
            
        return max_points

e3:T4d0,class Solution: def longestSubarray(self, nums: List[int], limit: int) -> int: max_d = deque() # Monotonically decreasing (front is max) min_d = deque() # Monotonically increasing (front is min) left = 0 res = 0 for right, num in enumerate(nums): # 1. Maintain max_d: remove elements smaller than current num while max_d and max_d[-1] < num: max_d.pop() max_d.append(num) # 2. Maintain min_d: remove elements larger than current num while min_d and min_d[-1] > num: min_d.pop() min_d.append(num) # 3. Check condition: if diff > limit, shrink window from left while max_d[0] - min_d[0] > limit: # If the number leaving the window was the max or min, # pop it from the respective deque if max_d[0] == nums[left]: max_d.popleft() if min_d[0] == nums[left]: min_d.popleft() left += 1 # 4. Update max length res = max(res, right - left + 1) return rese4:T3910,

class Solution:
    def longestSubarray(self, nums: List[int], limit: int) -> int:
        max_d = deque() # Monotonically decreasing (front is max)
        min_d = deque() # Monotonically increasing (front is min)
        left = 0
        res = 0
        
        for right, num in enumerate(nums):
            # 1. Maintain max_d: remove elements smaller than current num
            while max_d and max_d[-1] < num:
                max_d.pop()
            max_d.append(num)
            
            # 2. Maintain min_d: remove elements larger than current num
            while min_d and min_d[-1] > num:
                min_d.pop()
            min_d.append(num)
            
            # 3. Check condition: if diff > limit, shrink window from left
            while max_d[0] - min_d[0] > limit:
                # If the number leaving the window was the max or min, 
                # pop it from the respective deque
                if max_d[0] == nums[left]:
                    max_d.popleft()
                if min_d[0] == nums[left]:
                    min_d.popleft()
                left += 1
            
            # 4. Update max length
            res = max(res, right - left + 1)
            
        return res

e5:T4b1,Standard sliding window solutions (O(N)) only work when the array contains non-negative numbers. Because `nums` can contain negative numbers here, the prefix sum is not monotonic. This solution addresses that specific constraint efficiently. 1. **Prefix Sums:** We transform the problem to finding y and x (y > x) such that P[y] - P[x] >= K. 2. **Deque Logic:** * **Condition 1 (`popleft`):** Once an index x (from the front of the deque) satisfies the condition with current index y, we calculate the length and discard x. We never need x again because any future y' > y would result in a longer subarray (y' - x > y - x), and we are looking for the *shortest*. * **Condition 2 (`pop`):** If the current prefix sum P[y] is smaller than the prefix sum at the back of the deque (P[x]), we discard x. Why? Because y has a smaller value (making it easier to reach K) and y is a later index (making the subarray shorter). y is strictly better than x. ### Complexity Analysis * **Time Complexity:** O(N). Each index from 0 to N is added to the deque exactly once and popped from the deque at most once. * **Space Complexity:** O(N). Used for the Prefix Sum array and the Deque.e6:T576,

Standard sliding window solutions (O(N)) only work when the array contains non-negative numbers. Because nums can contain negative numbers here, the prefix sum is not monotonic. This solution addresses that specific constraint efficiently.

Prefix Sums: We transform the problem to finding y and x (y > x) such that P[y] - P[x] >= K.
Deque Logic:
- Condition 1 (popleft): Once an index x (from the front of the deque) satisfies the condition with current index y, we calculate the length and discard x. We never need x again because any future y' > y would result in a longer subarray (y' - x > y - x), and we are looking for the shortest.
- Condition 2 (pop): If the current prefix sum P[y] is smaller than the prefix sum at the back of the deque (P[x]), we discard x. Why? Because y has a smaller value (making it easier to reach K) and y is a later index (making the subarray shorter). y is strictly better than x.

Complexity Analysis

Time Complexity: O(N).
Each index from 0 to N is added to the deque exactly once and popped from the deque at most once.
Space Complexity: O(N).
Used for the Prefix Sum array and the Deque.

e7:T4ce,class Solution: def shortestSubarray(self, nums: List[int], k: int) -> int: n = len(nums) # Build Prefix Sum array # P[i] represents sum(nums[0]...nums[i-1]) P = [0] * (n + 1) for i in range(n): P[i + 1] = P[i] + nums[i] # Deque stores indices 'i' of the prefix sum array. # It is maintained to be monotonically increasing based on P[i]. dq = deque() min_len = n + 1 for y in range(n + 1): # 1. Check for valid subarrays ending at y # If P[y] - P[dq[0]] >= k, we found a valid subarray. # We try to shorten it by popping from the left. while dq and P[y] - P[dq[0]] >= k: min_len = min(min_len, y - dq.popleft()) # 2. Maintain monotonicity (Optimization) # If P[y] <= P[dq[-1]], then index y is a better start point # than dq[-1] because y is larger (shorter subarray) # and P[y] is smaller (easier to reach sum k). while dq and P[y] <= P[dq[-1]]: dq.pop() dq.append(y) return min_len if min_len <= n else -1e8:T3b15,

class Solution:
    def shortestSubarray(self, nums: List[int], k: int) -> int:
        n = len(nums)
        
        # Build Prefix Sum array
        # P[i] represents sum(nums[0]...nums[i-1])
        P = [0] * (n + 1)
        for i in range(n):
            P[i + 1] = P[i] + nums[i]
        
        # Deque stores indices 'i' of the prefix sum array.
        # It is maintained to be monotonically increasing based on P[i].
        dq = deque()
        min_len = n + 1
        
        for y in range(n + 1):
            # 1. Check for valid subarrays ending at y
            # If P[y] - P[dq[0]] >= k, we found a valid subarray.
            # We try to shorten it by popping from the left.
            while dq and P[y] - P[dq[0]] >= k:
                min_len = min(min_len, y - dq.popleft())
            
            # 2. Maintain monotonicity (Optimization)
            # If P[y] <= P[dq[-1]], then index y is a better start point
            # than dq[-1] because y is larger (shorter subarray) 
            # and P[y] is smaller (easier to reach sum k).
            while dq and P[y] <= P[dq[-1]]:
                dq.pop()
            
            dq.append(y)
            
        return min_len if min_len <= n else -1

e9:T2fec,

class Solution:
    def validPath(self, n: int, edges: List[List[int]], source: int, destination: int) -> bool:
        # Edge case: If start and end are the same, we are already there
        if source == destination:
            return True
        
        # 1. Build Adjacency List
        graph = defaultdict(list)
        for u, v in edges:
            graph[u].append(v)
            graph[v].append(u)
            
        # 2. Initialize BFS
        visited = set([source])
        queue = deque([source])
        
        # 3. Process Queue
        while queue:
            current_node = queue.popleft()
            
            # Optional: Check here (or before adding to queue)
            if current_node == destination:
                return True
            
            for neighbor in graph[current_node]:
                if neighbor not in visited:
                    visited.add(neighbor)
                    queue.append(neighbor)
                    
        return False

ea:T40d9,

class Solution:
    def validPath(self, n: int, edges: List[List[int]], source: int, destination: int) -> bool:
        if source == destination:
            return True
            
        root = list(range(n))
        rank = [1] * n
        
        def find(x):
            if root[x] != x:
                root[x] = find(root[x]) # Path compression
            return root[x]
        
        def union(x, y):
            rootX, rootY = find(x), find(y)
            if rootX != rootY:
                # Union by rank optimization
                if rank[rootX] > rank[rootY]:
                    root[rootY] = rootX
                elif rank[rootX] < rank[rootY]:
                    root[rootX] = rootY
                else:
                    root[rootY] = rootX
                    rank[rootX] += 1
        
        for u, v in edges:
            union(u, v)
            
        return find(source) == find(destination)

eb:T3727,

class Solution:
    def findAnagrams(self, s: str, p: str) -> list[int]:
        if len(p) > len(s):
            return []

        p_count = [0] * 26
        s_count = [0] * 26
        result = []
        
        # Calculate frequency for p and the initial window of s
        for char in p:
            p_count[ord(char) - ord('a')] += 1
            
        # Sliding window
        for i in range(len(s)):
            # Add the new character to the current window
            s_count[ord(s[i]) - ord('a')] += 1
            
            # If window size is greater than len(p), remove the leftmost character
            if i >= len(p):
                s_count[ord(s[i - len(p)]) - ord('a')] -= 1
            
            # Compare the frequency arrays
            if p_count == s_count:
                result.append(i - len(p) + 1)
                
        return result

ec:T2047,

class Solution:
    def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode':
        # Base Case:
        # 1. If the tree is empty, return None.
        # 2. If we found either p or q, return that node.
        if not root or root == p or root == q:
            return root

        # Recursive Step: Look for p and q in the left and right subtrees
        left = self.lowestCommonAncestor(root.left, p, q)
        right = self.lowestCommonAncestor(root.right, p, q)

        # Logic:
        # If both left and right returned a node, it means p is in one side 
        # and q is in the other. Therefore, the current 'root' is the split point (LCA).
        if left and right:
            return root
        
        # If only one side returned a node, it means both p and q are in that specific subtree 
        # (or one is inside the other). Return the non-None side.
        return left or right

ed:T49f,class Solution: def reorderList(self, head: ListNode) -> None: if not head or not head.next: return # Step 1: Find the Middle slow, fast = head, head while fast and fast.next: slow = slow.next fast = fast.next.next # 'slow' is now at the end of the first half. # We split the list here. second = slow.next slow.next = None # Cut the first half # Step 2: Reverse Second Half prev = None curr = second while curr: next_temp = curr.next curr.next = prev prev = curr curr = next_temp # 'prev' is now the head of the reversed second half # Step 3: Merge the Two Halves first, second = head, prev while second: # Save next pointers tmp1, tmp2 = first.next, second.next # Link first to second first.next = second # Link second to old first.next second.next = tmp1 # Move pointers forward first, second = tmp1, tmp2ee:T3081,

class Solution:
    def reorderList(self, head: ListNode) -> None:
        if not head or not head.next:
            return

        # Step 1: Find the Middle
        slow, fast = head, head
        while fast and fast.next:
            slow = slow.next
            fast = fast.next.next
        
        # 'slow' is now at the end of the first half.
        # We split the list here.
        second = slow.next
        slow.next = None # Cut the first half

        # Step 2: Reverse Second Half
        prev = None
        curr = second
        while curr:
            next_temp = curr.next
            curr.next = prev
            prev = curr
            curr = next_temp
        
        # 'prev' is now the head of the reversed second half
        
        # Step 3: Merge the Two Halves
        first, second = head, prev
        
        while second:
            # Save next pointers
            tmp1, tmp2 = first.next, second.next
            
            # Link first to second
            first.next = second
            # Link second to old first.next
            second.next = tmp1
            
            # Move pointers forward
            first, second = tmp1, tmp2

ef:T2384,

class Solution:
    def mergeTwoLists(self, list1: Optional[ListNode], list2: Optional[ListNode]) -> Optional[ListNode]:
        # 1. Create a dummy node to act as the pre-head of the result.
        # This avoids checking if the result head is None during the first iteration.
        dummy = ListNode()
        tail = dummy
        
        # 2. Iterate while both lists have nodes remaining
        while list1 and list2:
            if list1.val < list2.val:
                tail.next = list1
                list1 = list1.next
            else:
                tail.next = list2
                list2 = list2.next
            tail = tail.next
        
        # 3. Attach the remaining nodes.
        # Since the lists are sorted, we just attach the remainder of the non-empty list.
        # In Python, 'or' returns the first truthy value.
        tail.next = list1 or list2
        
        # 4. Return the next node after dummy (the actual head)
        return dummy.next

f0:T4b4,class Solution: def checkInclusion(self, s1: str, s2: str) -> bool: if len(s1) > len(s2): return False # Initialize frequency arrays for a-z s1_count = [0] * 26 s2_count = [0] * 26 # Fill frequency arrays for the first window for i in range(len(s1)): s1_count[ord(s1[i]) - ord('a')] += 1 s2_count[ord(s2[i]) - ord('a')] += 1 # Check the very first window if s1_count == s2_count: return True # Slide the window across s2 # i represents the index of the character entering the window for i in range(len(s1), len(s2)): # Add the new character to the window s2_count[ord(s2[i]) - ord('a')] += 1 # Remove the character that just left the window # The character leaving is at index: i - len(s1) left_char_idx = i - len(s1) s2_count[ord(s2[left_char_idx]) - ord('a')] -= 1 # Check if the current window matches s1 if s1_count == s2_count: return True return Falsef1:T40e2,

class Solution:
    def checkInclusion(self, s1: str, s2: str) -> bool:
        if len(s1) > len(s2):
            return False
        
        # Initialize frequency arrays for a-z
        s1_count = [0] * 26
        s2_count = [0] * 26
        
        # Fill frequency arrays for the first window
        for i in range(len(s1)):
            s1_count[ord(s1[i]) - ord('a')] += 1
            s2_count[ord(s2[i]) - ord('a')] += 1
            
        # Check the very first window
        if s1_count == s2_count:
            return True
            
        # Slide the window across s2
        # i represents the index of the character entering the window
        for i in range(len(s1), len(s2)):
            # Add the new character to the window
            s2_count[ord(s2[i]) - ord('a')] += 1
            
            # Remove the character that just left the window
            # The character leaving is at index: i - len(s1)
            left_char_idx = i - len(s1)
            s2_count[ord(s2[left_char_idx]) - ord('a')] -= 1
            
            # Check if the current window matches s1
            if s1_count == s2_count:
                return True
                
        return False

f2:T431,

This sliding window solution achieves O(N) time complexity with a subtle optimization: it never actually shrinks the window size; it only shifts it or allows it to grow.

The Key Logic (The "Trick")
You might notice that inside the if block (when the window is invalid), we do not decrement max_freq.

In a standard sliding window, when you remove a character from the left, the "most frequent character" might change, requiring a re-scan of your counts. However, here we are only looking for the longest substring.

If we shrink the window, the window size decreases. We don't care about valid windows smaller than the best one we've already found.
The window only expands (and updates the result) if we find a new max_freq that is higher than before, relative to the window size.
Therefore, max_freq effectively acts as a "historic maximum frequency" for the current window width. We don't need to lower it because the window width will not grow again until we find a frequency that exceeds that historic max.

f3:T2a2d,

class Solution:
    def characterReplacement(self, s: str, k: int) -> int:
        count = {}
        max_freq = 0
        left = 0
        
        for right in range(len(s)):
            # Add current character to the count
            char = s[right]
            count[char] = count.get(char, 0) + 1
            
            # Update the count of the most frequent character in the current window
            max_freq = max(max_freq, count[char])
            
            # Validity check:
            # window_length - max_freq = number of replacements needed
            # If replacements needed > k, the window is invalid.
            if (right - left + 1) - max_freq > k:
                # Remove the left character from the map and shift left pointer
                count[s[left]] -= 1
                left += 1
                
        # The window size is (right - left + 1). 
        # Since 'right' ended at len(s) - 1, the max length found is len(s) - left.
        return len(s) - left

f4:T2780,

class Solution:
    def searchMatrix(self, matrix: list[list[int]], target: int) -> bool:
        if not matrix:
            return False
        
        m, n = len(matrix), len(matrix[0])
        left, right = 0, m * n - 1
        
        while left <= right:
            mid = (left + right) // 2
            # Convert 1D index 'mid' to 2D coordinates
            row = mid // n
            col = mid % n
            guess = matrix[row][col]
            
            if guess == target:
                return True
            elif guess < target:
                left = mid + 1
            else:
                right = mid - 1
                
        return False

f5:T1da9,

class Solution:
    def findMin(self, nums: List[int]) -> int:
        left, right = 0, len(nums) - 1
        
        while left < right:
            mid = (left + right) // 2
            
            # If the middle element is greater than the rightmost element,
            # the minimum value must be in the right half.
            if nums[mid] > nums[right]:
                left = mid + 1
            
            # Otherwise, the minimum is either at mid or in the left half.
            # We keep 'mid' in the search space.
            else:
                right = mid
                
        # When the loop terminates, left == right, pointing to the minimum.
        return nums[left]

f6:T2dc9,

class Solution:
    def topKFrequent(self, nums: List[int], k: int) -> List[int]:
        count = {}
        # Create buckets where index = frequency
        # Size is len(nums) + 1 because max frequency is len(nums)
        freq = [[] for _ in range(len(nums) + 1)]
        
        # 1. Count frequencies
        for n in nums:
            count[n] = count.get(n, 0) + 1
            
        # 2. Fill buckets: index is frequency, value is list of numbers
        for n, c in count.items():
            freq[c].append(n)
            
        # 3. Iterate backwards (highest frequency first) to collect k elements
        res = []
        for i in range(len(freq) - 1, 0, -1):
            for n in freq[i]:
                res.append(n)
                if len(res) == k:
                    return res
        return res

f7:T145a,

from collections import Counter
import heapq
from typing import List

def topKFrequent(nums: List[int], k: int) -> List[int]:
    # 1. Build the frequency map: O(N)
    count = Counter(nums)
    
    # 2. Use a heap to find the k largest frequent elements: O(N log k)
    # heapq.nlargest optimizes this process internally
    return heapq.nlargest(k, count.keys(), key=count.get)

28:["$","$L39",null,{"children":["$","$L3a",null,{"initialItems":[{"id":"734f7890-332f-4196-bc14-7656cff32092","userId":"$undefined","title":"How do you know when a research rabbit hole is still useful?","slug":"research-benders-need-stop-rules","url":"https://github.com/teamleaderleo/linux-fieldwork/blob/c79d34b65fb2cd8c54234f361f073248c53b513a/notes/research-benders-need-discriminators-and-stop-rules.md","defaultIndex":0,"versions":[{"label":"Read","content":"$3b","contentHtml":"$3c","code":null,"codeHtml":""},{"label":"Try it","content":"## Try it\n\nTake one open technical question and write four lines:\n\n1. **Decision:** what choice could this work change?\n2. **Discriminator:** what result would make the leading explanation lose?\n3. **Stop rule:** when is the current question saturated?\n4. **Reopen trigger:** what new evidence would make another pass worthwhile?\n\nIf line 2 is impossible to write, the question is probably still too vague.","contentHtml":"

Try it

Take one open technical question and write four lines:

Decision: what choice could this work change?
Discriminator: what result would make the leading explanation lose?
Stop rule: when is the current question saturated?
Reopen trigger: what new evidence would make another pass worthwhile?

If line 2 is impossible to write, the question is probably still too vague.

","code":null,"codeHtml":""}],"tags":["source:linux-fieldwork","collection:fieldwork-studies-01","mode:review","domain:research","time:5-min","device:phone","state:fresh","visibility:public"],"category":"review","createdAt":1786293421380,"updatedAt":1786293421380,"score":100,"review":"$undefined"},{"id":"c86b76f2-26c9-4931-9950-f0796f6175b9","userId":"$undefined","title":"Why should a cache filename mean “complete object”?","slug":"cache-files-are-published-atomically","url":"https://github.com/teamleaderleo/linux-fieldwork/blob/c79d34b65fb2cd8c54234f361f073248c53b513a/notes/reliability/cache-files-must-be-published-atomically.md","defaultIndex":0,"versions":[{"label":"Read","content":"## The question\n\nWhat can go wrong when a threaded cache writes a response directly to its final filename?\n\n## Working model\n\nThe final name becomes visible as soon as the writer opens it. A second request can mistake “a writer has started” for “the object is complete” and return the partial bytes as a successful cache hit.\n\nWrite to a unique temporary file in the destination directory, validate the whole object, close the writer, and atomically replace the final path. Remove the temporary file on every failure. Same-directory publication keeps the rename on one filesystem.\n\nAtomic publication solves visibility, not every race. Two misses may still download twice, and integrity checks such as expected length remain separate.\n\n## Provenance\n\nSynthesized from Linux Fieldwork's mmdebstrap cache-proxy investigation at revision `c79d34b65fb2cd8c54234f361f073248c53b513a`.","contentHtml":"

The question

What can go wrong when a threaded cache writes a response directly to its final filename?

Working model

The final name becomes visible as soon as the writer opens it. A second request can mistake “a writer has started” for “the object is complete” and return the partial bytes as a successful cache hit.

Write to a unique temporary file in the destination directory, validate the whole object, close the writer, and atomically replace the final path. Remove the temporary file on every failure. Same-directory publication keeps the rename on one filesystem.

Atomic publication solves visibility, not every race. Two misses may still download twice, and integrity checks such as expected length remain separate.

Provenance

Synthesized from Linux Fieldwork's mmdebstrap cache-proxy investigation at revision c79d34b65fb2cd8c54234f361f073248c53b513a.

","code":null,"codeHtml":""},{"label":"Try it","content":"## Try it\n\nSketch a two-request timeline. Request A creates the final path and pauses after one chunk. Request B arrives.\n\nThen replace the direct write with `temporary → validate → close → atomic rename`. Which states can B observe before and after the change? What extra mechanism would you need to prevent duplicate downloads, and why is that a different requirement?","contentHtml":"

Try it

Sketch a two-request timeline. Request A creates the final path and pauses after one chunk. Request B arrives.

Then replace the direct write with temporary → validate → close → atomic rename. Which states can B observe before and after the change? What extra mechanism would you need to prevent duplicate downloads, and why is that a different requirement?

","code":null,"codeHtml":""}],"tags":["source:linux-fieldwork","collection:fieldwork-studies-01","mode:explain","domain:reliability","time:5-min","device:phone","state:fresh","visibility:public"],"category":"explain","createdAt":1786293421380,"updatedAt":1786293421380,"score":100,"review":"$undefined"},{"id":"1a8a4743-54e3-4b0e-84da-836fa76cac03","userId":"$undefined","title":"Why does Python return −15 for a terminated child?","slug":"negative-subprocess-codes-are-signals","url":"https://github.com/teamleaderleo/linux-fieldwork/blob/c79d34b65fb2cd8c54234f361f073248c53b513a/notes/processes/negative-subprocess-returncodes-are-signals.md","defaultIndex":0,"versions":[{"label":"Read","content":"## The question\n\nWhy should a Python wrapper never pass a child's negative `subprocess` return code straight to `sys.exit()`?\n\n## Working model\n\nIn Python, a negative return code means the child died from a signal: `-15` means SIGTERM. `SystemExit(-15)` creates an ordinary exit whose low byte is 241. That loses both signal identity and the conventional shell status 143.\n\nA wrapper must choose deliberately: restore the default handler and re-signal itself to preserve exact signal termination, or map to `128 + signal` for a normal shell-style exit. Finish required local cleanup before re-signaling.\n\nPositive codes remain ordinary exit statuses.\n\n## Provenance\n\nSynthesized from Linux Fieldwork's proxysolver signal-status investigation at revision `c79d34b65fb2cd8c54234f361f073248c53b513a`.","contentHtml":"

The question

Why should a Python wrapper never pass a child's negative subprocess return code straight to sys.exit()?

Working model

In Python, a negative return code means the child died from a signal: -15 means SIGTERM. SystemExit(-15) creates an ordinary exit whose low byte is 241. That loses both signal identity and the conventional shell status 143.

A wrapper must choose deliberately: restore the default handler and re-signal itself to preserve exact signal termination, or map to 128 + signal for a normal shell-style exit. Finish required local cleanup before re-signaling.

Positive codes remain ordinary exit statuses.

Provenance

Synthesized from Linux Fieldwork's proxysolver signal-status investigation at revision c79d34b65fb2cd8c54234f361f073248c53b513a.

","code":null,"codeHtml":""},{"label":"Try it","content":"## Predict the observer\n\nFor child results `0`, `7`, and `-15`, write what these observers should see:\n\n- another Python parent;\n- a POSIX shell;\n- `WIFSIGNALED` / `WTERMSIG`.\n\nCompare exact re-raising with mapping SIGTERM to 143. Which contract does your wrapper need?","contentHtml":"

Predict the observer

For child results 0, 7, and -15, write what these observers should see:

another Python parent;
a POSIX shell;
WIFSIGNALED / WTERMSIG.

Compare exact re-raising with mapping SIGTERM to 143. Which contract does your wrapper need?

","code":"if returncode < 0:\n signum = -returncode\n # re-raise exactly, or map to 128 + signum\nelif returncode > 0:\n raise SystemExit(returncode)","codeHtml":""}],"tags":["source:linux-fieldwork","collection:fieldwork-studies-01","mode:explain","domain:python","time:5-min","device:phone","state:fresh","visibility:public"],"category":"explain","createdAt":1786293421380,"updatedAt":1786293421380,"score":100,"review":"$undefined"},{"id":"a66bd307-6170-4442-b905-db0b1e2f7535","userId":"$undefined","title":"Why is file replacement really two state machines?","slug":"install-is-two-state-machines","url":"https://github.com/teamleaderleo/linux-fieldwork/blob/c79d34b65fb2cd8c54234f361f073248c53b513a/notes/filesystems/coreutils-install-state-machines.md","defaultIndex":0,"versions":[{"label":"Read","content":"$3d","contentHtml":"$3e","code":null,"codeHtml":""},{"label":"Try it","content":"## Trace it\n\nStart with `dest/file = original`. Source A fails during data copy; source B later succeeds to the same name.\n\nDraw both machines separately and mark these events:\n\n- old destination renamed to backup;\n- A begins, then fails;\n- partial A removed;\n- backup restored;\n- B copies successfully.\n\nThe expected end state is `dest/file = B` and `dest/file~ = original`. Explain why marking A as an owner would produce the wrong answer.","contentHtml":"

Trace it

Start with dest/file = original. Source A fails during data copy; source B later succeeds to the same name.

Draw both machines separately and mark these events:

old destination renamed to backup;
A begins, then fails;
partial A removed;
backup restored;
B copies successfully.

The expected end state is dest/file = B and dest/file~ = original. Explain why marking A as an owner would produce the wrong answer.

","code":null,"codeHtml":""}],"tags":["source:linux-fieldwork","collection:fieldwork-studies-01","mode:trace","domain:filesystems","time:20-min","device:phone","state:fresh","visibility:public","prep:interview"],"category":"trace","createdAt":1786293421380,"updatedAt":1786293421380,"score":100,"review":"$undefined"},{"id":"f5d34eed-84a6-4d72-98ce-4947b5e9376b","userId":"$undefined","title":"Why must a proxy rewrite headers after decoding a body?","slug":"proxies-normalize-framing-after-decoding","url":"https://github.com/teamleaderleo/linux-fieldwork/blob/c79d34b65fb2cd8c54234f361f073248c53b513a/notes/reliability/proxies-must-normalize-framing-after-decoding.md","defaultIndex":0,"versions":[{"label":"Read","content":"## The question\n\nWhat happens if an HTTP library removes chunk framing from a body, but a proxy forwards `Transfer-Encoding: chunked` unchanged?\n\n## Working model\n\nChunked encoding describes wire bytes. A client library may consume chunk sizes, separators, the terminal chunk, and trailers before returning only payload bytes. If the proxy sends those decoded bytes while advertising chunked encoding, the next client tries to parse payload text as chunk sizes.\n\nHeaders must describe the representation actually sent on the next hop. Remove stale transfer framing and hop-by-hop fields, including fields named by `Connection`, then choose a valid downstream boundary: correct length, newly generated chunk framing, or close delimiting.\n\nA correct cache file does not prove the first streamed response was valid.\n\n## Provenance\n\nSynthesized from a loopback regression against Python's HTTP stack at Linux Fieldwork revision `c79d34b65fb2cd8c54234f361f073248c53b513a`.","contentHtml":"$3f","code":null,"codeHtml":""},{"label":"Try it","content":"## Trace the representations\n\nLabel each line as **wire framing** or **logical payload**. Then write the valid downstream headers for a seven-byte payload.\n\nThe trap is preserving metadata from representation A after a library has already transformed the bytes into representation B.","contentHtml":"

Trace the representations

Label each line as wire framing or logical payload. Then write the valid downstream headers for a seven-byte payload.

The trap is preserving metadata from representation A after a library has already transformed the bytes into representation B.

","code":"HTTP/1.1 200 OK\nTransfer-Encoding: chunked\n\n7\npayload\n0\n\n\n# decoded body returned by the client library\npayload","codeHtml":""}],"tags":["source:linux-fieldwork","collection:fieldwork-studies-01","mode:trace","domain:http","time:20-min","device:phone","state:fresh","visibility:public","prep:interview"],"category":"trace","createdAt":1786293421380,"updatedAt":1786293421380,"score":100,"review":"$undefined"},{"id":"4cd09fba-2561-4091-a5ea-17e390c525e1","userId":"$undefined","title":"Should a cleanup-time signal replace an already completed failure?","slug":"completed-failures-outlive-cleanup-signals","url":"https://github.com/teamleaderleo/linux-fieldwork/blob/c79d34b65fb2cd8c54234f361f073248c53b513a/notes/processes/completed-results-must-not-be-replaced-by-later-cleanup-signals.md","defaultIndex":0,"versions":[{"label":"Read","content":"## The question\n\nA child durably reports failure, host cleanup begins, and then SIGTERM arrives. Which result should the wrapper report?\n\n## Working model\n\nPrecedence should follow event order and ownership, not the order in which cleanup happens to inspect variables. If the child result was complete and durable before cleanup began, a later cleanup-time signal matters after success but should not replace that earlier authoritative failure.\n\nA useful order for this proven sequence is: captured host failure; completed child failure; first cleanup-time signal; first cleanup failure; success.\n\nThis is not universal. If the child result was still provisional or buffered, the signal may be the first authoritative result.\n\n## Provenance\n\nSynthesized from Linux Fieldwork result-precedence investigations at revision `c79d34b65fb2cd8c54234f361f073248c53b513a`.","contentHtml":"

The question

A child durably reports failure, host cleanup begins, and then SIGTERM arrives. Which result should the wrapper report?

Working model

Precedence should follow event order and ownership, not the order in which cleanup happens to inspect variables. If the child result was complete and durable before cleanup began, a later cleanup-time signal matters after success but should not replace that earlier authoritative failure.

A useful order for this proven sequence is: captured host failure; completed child failure; first cleanup-time signal; first cleanup failure; success.

This is not universal. If the child result was still provisional or buffered, the signal may be the first authoritative result.

Provenance

Synthesized from Linux Fieldwork result-precedence investigations at revision c79d34b65fb2cd8c54234f361f073248c53b513a.

","code":null,"codeHtml":""},{"label":"Try it","content":"## Build the table\n\nFill in the final result for:\n\n- child success → cleanup-time SIGINT;\n- child failure → cleanup-time SIGINT;\n- missing child result → cleanup-time SIGINT;\n- host failure → child failure → cleanup-time SIGINT.\n\nFor every answer, name when the winning result became final and who owned it.","contentHtml":"

Build the table

Fill in the final result for:

child success → cleanup-time SIGINT;
child failure → cleanup-time SIGINT;
missing child result → cleanup-time SIGINT;
host failure → child failure → cleanup-time SIGINT.

For every answer, name when the winning result became final and who owned it.

","code":null,"codeHtml":""}],"tags":["source:linux-fieldwork","collection:fieldwork-studies-01","mode:review","domain:processes","time:20-min","device:phone","state:fresh","visibility:public","prep:interview"],"category":"review","createdAt":1786293421380,"updatedAt":1786293421380,"score":100,"review":"$undefined"},{"id":"f55bbbd8-003c-4938-9045-c6ffa6724b94","userId":"$undefined","title":"How do you keep EXIT cleanup from rewriting the real result?","slug":"result-precedence-survives-exit-cleanup","url":"https://github.com/teamleaderleo/linux-fieldwork/blob/c79d34b65fb2cd8c54234f361f073248c53b513a/notes/processes/result-precedence-must-survive-exit-cleanup.md","defaultIndex":0,"versions":[{"label":"Read","content":"## The question\n\nA shell wrapper can observe command failure, subordinate failure, a signal, and cleanup failure. Why does “last error wins” describe control flow rather than meaning?\n\n## Working model\n\nCapture the primary result before cleanup and encode precedence explicitly. A primary command or first handled-signal failure outranks a subordinate status channel; that outranks the first cleanup failure; success comes last.\n\nDo not share one `$?`-based trap body across `EXIT`, `INT`, and `TERM`. A signal may be deferred while the shell waits, so `$?` can describe the completed child instead of the signal. Give signal handlers explicit statuses, prevent EXIT from running cleanup twice, and stabilize the first handled result while cleanup continues.\n\nFallible result-file reads also need containment so `set -e` cannot manufacture a new final failure.\n\n## Provenance\n\nSynthesized from Linux Fieldwork wrapper regressions at revision `c79d34b65fb2cd8c54234f361f073248c53b513a`.","contentHtml":"$40","code":null,"codeHtml":""},{"label":"Try it","content":"## Trace the traps\n\nRun the shared-trap example mentally for a deferred TERM while the shell waits for a foreground child. What can `$?` contain when the handler finally runs?\n\nThen rewrite the policy as a small precedence function. Keep cleanup once-only and retain the first cleanup failure while still attempting later cleanup.","contentHtml":"

Trace the traps

Run the shared-trap example mentally for a deferred TERM while the shell waits for a foreground child. What can $? contain when the handler finally runs?

Then rewrite the policy as a small precedence function. Keep cleanup once-only and retain the first cleanup failure while still attempting later cleanup.

","code":"trap cleanup INT TERM EXIT\n\ncleanup() {\n status=$?\n # Which event does status actually describe?\n finish \"$status\"\n}","codeHtml":""}],"tags":["source:linux-fieldwork","collection:fieldwork-studies-01","mode:trace","domain:shell","time:20-min","device:phone","state:fresh","visibility:public","prep:interview"],"category":"trace","createdAt":1786293421380,"updatedAt":1786293421380,"score":100,"review":"$undefined"},{"id":"e96b21d7-cea4-4302-9d6a-bb9f6ce63eb3","userId":"$undefined","title":"Why must a proxy separate proxy and origin credentials?","slug":"proxy-and-origin-credentials-are-different","url":"https://github.com/teamleaderleo/linux-fieldwork/blob/c79d34b65fb2cd8c54234f361f073248c53b513a/notes/security/proxies-must-separate-proxy-and-origin-credentials.md","defaultIndex":0,"versions":[{"label":"Read","content":"## The question\n\nWhy is forwarding `dict(request.headers)` from a proxy to an origin both a security bug and a compatibility bug?\n\n## Working model\n\nHeaders have scope. `Proxy-Authorization` is a credential for the proxy, not a server selected by the request URL. Connection controls apply to one transport hop. `Connection` can name additional hop-specific fields, and header names are case-insensitive.\n\nConstruct the origin request from an explicit policy: remove proxy credentials and hop-by-hop fields, parse every `Connection` token, preserve one valid Host, and retain safe end-to-end fields—including repeated fields when their meaning requires it.\n\nOrigin `Authorization` is different from `Proxy-Authorization`; deleting both is not a safe shortcut.\n\n## Provenance\n\nSynthesized from a loopback mmdebstrap proxy regression at Linux Fieldwork revision `c79d34b65fb2cd8c54234f361f073248c53b513a`. It does not define every extension or CONNECT policy.","contentHtml":"$41","code":null,"codeHtml":""},{"label":"Try it","content":"## Review the request\n\nDecide what reaches the origin:\n\n- `Proxy-Authorization: Basic fake-secret`\n- `Authorization: Bearer origin-token`\n- `Connection: keep-alive, X-Hop`\n- `X-Hop: local-only`\n- `Range: bytes=0-1023`\n\nThen state what a real loopback-origin test can prove that a helper-unit test cannot.","contentHtml":"

Review the request

Decide what reaches the origin:

Proxy-Authorization: Basic fake-secret
Authorization: Bearer origin-token
Connection: keep-alive, X-Hop
X-Hop: local-only
Range: bytes=0-1023

Then state what a real loopback-origin test can prove that a helper-unit test cannot.

","code":null,"codeHtml":""}],"tags":["source:linux-fieldwork","collection:fieldwork-studies-01","mode:review","domain:security","time:20-min","device:phone","state:fresh","visibility:public","prep:interview"],"category":"review","createdAt":1786293421380,"updatedAt":1786293421380,"score":100,"review":"$undefined"},{"id":"8fb759e7-d906-44bd-8679-e5b174aebf3d","userId":"$undefined","title":"What changes when you model ownership before proposing a patch?","slug":"build-the-ownership-model-before-the-patch","url":"https://github.com/teamleaderleo/fieldwork/blob/49e8c7dabbc75f021b09d243b1d95ecac215bb07/research/vite-process-notes-2026-08-09.md","defaultIndex":0,"versions":[{"label":"Read","content":"## The question\n\nWhy is “this code looks suspicious” a weaker starting point than an observed lifecycle anomaly plus an ownership model?\n\n## Working model\n\nBegin with the resource or state: who created it, which lifetime owns it, what transfers or ends that ownership, what completion means, and what happens on success, error, cancellation, restart, warm state, and cold state.\n\nThen require a target-native regression that fails on a clean baseline before promoting a candidate. A plausible source theory is a hypothesis; a discriminator makes it a finding. Invalid probes are useful when you record how setup answered the wrong question.\n\nUse one finding to inspect adjacent branches, but make each branch earn promotion independently.\n\n## Provenance\n\nSynthesized from the Vite correctness-lane process notes at Fieldwork revision `49e8c7dabbc75f021b09d243b1d95ecac215bb07`. These are process observations, not Vite requirements.","contentHtml":"

The question

Why is “this code looks suspicious” a weaker starting point than an observed lifecycle anomaly plus an ownership model?

Working model

Begin with the resource or state: who created it, which lifetime owns it, what transfers or ends that ownership, what completion means, and what happens on success, error, cancellation, restart, warm state, and cold state.

Then require a target-native regression that fails on a clean baseline before promoting a candidate. A plausible source theory is a hypothesis; a discriminator makes it a finding. Invalid probes are useful when you record how setup answered the wrong question.

Use one finding to inspect adjacent branches, but make each branch earn promotion independently.

Provenance

Synthesized from the Vite correctness-lane process notes at Fieldwork revision 49e8c7dabbc75f021b09d243b1d95ecac215bb07. These are process observations, not Vite requirements.

","code":null,"codeHtml":""},{"label":"Try it","content":"## Apply the model\n\nPick one object from a running development server—a watcher, plugin container, dependency cache, child process, or bundle—and answer:\n\n1. Who creates it?\n2. What generation owns it?\n3. What operation ends that generation?\n4. What does settled cleanup mean?\n5. Which warm/cold or success/failure pair could make the theory lose?\n\nOnly then sketch the smallest regression.","contentHtml":"

Apply the model

Pick one object from a running development server—a watcher, plugin container, dependency cache, child process, or bundle—and answer:

Who creates it?
What generation owns it?
What operation ends that generation?
What does settled cleanup mean?
Which warm/cold or success/failure pair could make the theory lose?

Only then sketch the smallest regression.

","code":null,"codeHtml":""}],"tags":["source:fieldwork","collection:fieldwork-studies-01","mode:review","domain:javascript","time:5-min","device:phone","state:fresh","visibility:public","prep:interview"],"category":"review","createdAt":1786293421380,"updatedAt":1786293421380,"score":100,"review":"$undefined"},{"id":"c4460b4d-a33e-4872-a1e9-146d16b09c77","userId":"$undefined","title":"Why is end-of-stream not proof that a cache fill succeeded?","slug":"cache-fills-verify-declared-length","url":"https://github.com/teamleaderleo/linux-fieldwork/blob/c79d34b65fb2cd8c54234f361f073248c53b513a/notes/reliability/cache-fills-must-verify-declared-length.md","defaultIndex":0,"versions":[{"label":"Read","content":"## The question\n\nAn upstream promises 100 bytes, sends 40, and closes. Why can a cache still publish the short file even when the first client reports an incomplete response?\n\n## Working model\n\nSome HTTP read APIs return an ordinary empty read at premature EOF. The downstream client notices that the forwarded `Content-Length` was not met, but the proxy's loop may interpret EOF as success.\n\nWhen an upstream declares a nonnegative length, count the bytes written and compare the count before publication. On mismatch, fail the fill and let temporary-file cleanup remove the candidate. A later request can then reach the upstream again.\n\nAtomic rename and length validation compose: either one alone can still publish the wrong state.\n\n## Provenance\n\nSynthesized from a Python `http.client` / mmdebstrap investigation at Linux Fieldwork revision `c79d34b65fb2cd8c54234f361f073248c53b513a`.","contentHtml":"

The question

An upstream promises 100 bytes, sends 40, and closes. Why can a cache still publish the short file even when the first client reports an incomplete response?

Working model

Some HTTP read APIs return an ordinary empty read at premature EOF. The downstream client notices that the forwarded Content-Length was not met, but the proxy's loop may interpret EOF as success.

When an upstream declares a nonnegative length, count the bytes written and compare the count before publication. On mismatch, fail the fill and let temporary-file cleanup remove the candidate. A later request can then reach the upstream again.

Atomic rename and length validation compose: either one alone can still publish the wrong state.

Provenance

Synthesized from a Python http.client / mmdebstrap investigation at Linux Fieldwork revision c79d34b65fb2cd8c54234f361f073248c53b513a.

","code":null,"codeHtml":""},{"label":"Try it","content":"## Diagnose it\n\nFor each case, decide whether EOF is enough to publish:\n\n1. `Content-Length: 100`, received 100.\n2. `Content-Length: 100`, received 40.\n3. No declared length, response is validly close-delimited.\n\nThen explain why “the first client already failed” does not protect the next client from a poisoned cache.","contentHtml":"

Diagnose it

For each case, decide whether EOF is enough to publish:

Content-Length: 100, received 100.
Content-Length: 100, received 40.
No declared length, response is validly close-delimited.

Then explain why “the first client already failed” does not protect the next client from a poisoned cache.

","code":null,"codeHtml":""}],"tags":["source:linux-fieldwork","collection:fieldwork-studies-01","mode:debug","domain:http","time:5-min","device:phone","state:fresh","visibility:public","prep:interview"],"category":"debug","createdAt":1786293421380,"updatedAt":1786293421380,"score":100,"review":"$undefined"},{"id":"fea1c2ef-f385-424a-8fd3-5d7bab5d2bc6","userId":"$undefined","title":"Why can perfect cleanup still report a false success?","slug":"cancellation-cleanup-is-not-success","url":"https://github.com/teamleaderleo/linux-fieldwork/blob/c79d34b65fb2cd8c54234f361f073248c53b513a/notes/processes/cancellation-cleanup-must-not-fall-through-to-success.md","defaultIndex":0,"versions":[{"label":"Read","content":"## The question\n\nA wrapper catches Ctrl-C, terminates its child, waits, removes every temporary file, and then exits 0. What is wrong if nothing leaked?\n\n## Working model\n\nCleanup answers “did we leave the machine in a recoverable state?” Completion answers “did the requested operation finish?” They are different contracts.\n\nA handler that breaks or returns into the ordinary success epilogue can silently skip remaining work while CI reports green. For each handled signal, decide descendant delivery, waiting and escalation, resource cleanup, final status or signal identity, and a diagnostic that distinguishes cancellation from failure.\n\nExit 130 is a conventional handled-SIGINT result; exact signal re-raising is another valid contract. Accidental status 0 is not.\n\n## Provenance\n\nSynthesized from Linux Fieldwork's parent-only SIGINT regression at revision `c79d34b65fb2cd8c54234f361f073248c53b513a`.","contentHtml":"

The question

A wrapper catches Ctrl-C, terminates its child, waits, removes every temporary file, and then exits 0. What is wrong if nothing leaked?

Working model

Cleanup answers “did we leave the machine in a recoverable state?” Completion answers “did the requested operation finish?” They are different contracts.

A handler that breaks or returns into the ordinary success epilogue can silently skip remaining work while CI reports green. For each handled signal, decide descendant delivery, waiting and escalation, resource cleanup, final status or signal identity, and a diagnostic that distinguishes cancellation from failure.

Exit 130 is a conventional handled-SIGINT result; exact signal re-raising is another valid contract. Accidental status 0 is not.

Provenance

Synthesized from Linux Fieldwork's parent-only SIGINT regression at revision c79d34b65fb2cd8c54234f361f073248c53b513a.

","code":null,"codeHtml":""},{"label":"Try it","content":"## Review the handler\n\nFind the control-flow error. Then list assertions for status, surviving descendants, temporary state, a success marker, and an immediate clean rerun.\n\nBonus: why should the matrix include a signal sent only to the parent PID rather than only process-group delivery?","contentHtml":"

Review the handler

Find the control-flow error. Then list assertions for status, surviving descendants, temporary state, a success marker, and an immediate clean rerun.

Bonus: why should the matrix include a signal sent only to the parent PID rather than only process-group delivery?

","code":"try:\n child.wait()\nexcept KeyboardInterrupt:\n child.terminate()\n child.wait()\n break\n\n# ordinary success epilogue\nreturn 0","codeHtml":""}],"tags":["source:linux-fieldwork","collection:fieldwork-studies-01","mode:debug","domain:processes","time:5-min","device:phone","state:fresh","visibility:public","prep:interview"],"category":"debug","createdAt":1786293421380,"updatedAt":1786293421380,"score":100,"review":"$undefined"},{"id":"b21f6c93-9d3e-407d-a7c7-442f90bb7d1f","userId":"$undefined","title":"Why is `grep $user /etc/subuid` the wrong identity check?","slug":"subordinate-id-files-need-exact-account-matching","url":"https://github.com/teamleaderleo/linux-fieldwork/blob/c79d34b65fb2cd8c54234f361f073248c53b513a/notes/security/subordinate-id-files-require-exact-account-field-matching.md","defaultIndex":0,"versions":[{"label":"Read","content":"## The question\n\nHow can a fuzzy search in `/etc/subuid` make setup silently skip the user it was supposed to configure?\n\n## Working model\n\nSubordinate-ID files are colon-delimited records; account identity is field 1. An unanchored regex can match a different account containing the requested name, and punctuation in the name can become regex syntax.\n\nExtract field 1 and compare it as a fixed, whole string. Distinguish an absent file, an empty file, a delimiter-free malformed line, and an exact record. Apply the same semantics to subuid and subgid, and verify an immediate rerun does not append duplicates.\n\nThis check answers only whether an account has a record. Range overlap and allocation policy are separate contracts.\n\n## Provenance\n\nSynthesized from Linux Fieldwork's mmdebstrap package-test investigation at revision `c79d34b65fb2cd8c54234f361f073248c53b513a`.","contentHtml":"

The question

How can a fuzzy search in /etc/subuid make setup silently skip the user it was supposed to configure?

Working model

Subordinate-ID files are colon-delimited records; account identity is field 1. An unanchored regex can match a different account containing the requested name, and punctuation in the name can become regex syntax.

Extract field 1 and compare it as a fixed, whole string. Distinguish an absent file, an empty file, a delimiter-free malformed line, and an exact record. Apply the same semantics to subuid and subgid, and verify an immediate rerun does not append duplicates.

This check answers only whether an account has a record. Range overlap and allocation policy are separate contracts.

Provenance

Synthesized from Linux Fieldwork's mmdebstrap package-test investigation at revision c79d34b65fb2cd8c54234f361f073248c53b513a.

","code":null,"codeHtml":""},{"label":"Try it","content":"## Find the false positive\n\nRequested account: `debci`.\n\nExplain why `old-debci-helper:200000:65536` must not count. Then unpack what each flag protects in:\n\n`cut -s -d: -f1 /etc/subuid | grep -Fxq -- \"$user\"`\n\nAdd one test for a regex-significant username and one for a malformed delimiter-free line.","contentHtml":"

Find the false positive

Requested account: debci.

Explain why old-debci-helper:200000:65536 must not count. Then unpack what each flag protects in:

cut -s -d: -f1 /etc/subuid | grep -Fxq -- \"$user\"

Add one test for a regex-significant username and one for a malformed delimiter-free line.

","code":"cut -s -d: -f1 /etc/subuid | grep -Fxq -- \"$user\"","codeHtml":""}],"tags":["source:linux-fieldwork","collection:fieldwork-studies-01","mode:debug","domain:linux","time:5-min","device:phone","state:fresh","visibility:public","prep:interview"],"category":"debug","createdAt":1786293421380,"updatedAt":1786293421380,"score":100,"review":"$undefined"},{"id":"38a6cb20-53c4-4f74-8714-046f67ada52e","userId":"$undefined","title":"Why are real anomalies better research seeds than suspicious-looking code?","slug":"observed-friction-is-a-better-research-seed","url":"https://github.com/teamleaderleo/fieldwork/blob/49e8c7dabbc75f021b09d243b1d95ecac215bb07/research/vite-process-notes-2026-08-09.md","defaultIndex":0,"versions":[{"label":"Read","content":"## The question\n\nWhat makes an unexpected rebuild, a close that never settles, or state that changes across restart a strong starting point for source research?\n\n## Working model\n\nObserved friction already supplies a consequence and a boundary. Trace the anomaly back to the component that owns the result, then build a small native probe with a clean baseline failure. This is more informative than scanning for code that merely resembles a familiar defect.\n\nClassify failures along the way. A product regression, invalid fixture, discovery bug, platform flake, and formatting failure do not ask for the same repair. The first incomplete step has an owner.\n\nExploration can be messy; the eventual review shape should be one minimal change, its native regression, and the governing invariant.\n\n## Provenance\n\nSynthesized from the Vite correctness-lane process notes at Fieldwork revision `49e8c7dabbc75f021b09d243b1d95ecac215bb07`.","contentHtml":"

The question

What makes an unexpected rebuild, a close that never settles, or state that changes across restart a strong starting point for source research?

Working model

Observed friction already supplies a consequence and a boundary. Trace the anomaly back to the component that owns the result, then build a small native probe with a clean baseline failure. This is more informative than scanning for code that merely resembles a familiar defect.

Classify failures along the way. A product regression, invalid fixture, discovery bug, platform flake, and formatting failure do not ask for the same repair. The first incomplete step has an owner.

Exploration can be messy; the eventual review shape should be one minimal change, its native regression, and the governing invariant.

Provenance

Synthesized from the Vite correctness-lane process notes at Fieldwork revision 49e8c7dabbc75f021b09d243b1d95ecac215bb07.

","code":null,"codeHtml":""},{"label":"Try it","content":"## Turn friction into a probe\n\nTake one recent “that was weird” moment and write:\n\n- the first observable mismatch;\n- the smallest likely result owner;\n- a baseline behavior that should fail;\n- a control that proves the probe can distinguish the mechanism;\n- a stop condition if the anomaly belongs to the fixture or environment instead.\n\nDo not propose the patch yet.","contentHtml":"

Turn friction into a probe

Take one recent “that was weird” moment and write:

the first observable mismatch;
the smallest likely result owner;
a baseline behavior that should fail;
a control that proves the probe can distinguish the mechanism;
a stop condition if the anomaly belongs to the fixture or environment instead.

Do not propose the patch yet.

","code":null,"codeHtml":""}],"tags":["source:fieldwork","collection:fieldwork-studies-01","mode:debug","domain:research","time:5-min","device:phone","state:fresh","visibility:public","prep:interview"],"category":"debug","createdAt":1786293421380,"updatedAt":1786293421380,"score":100,"review":"$undefined"},{"id":"894ac452-266a-4b72-bed3-26c9105e0795","userId":"$undefined","title":"Stripe Practice: Fraudulent Transaction Detection","slug":"stripe-fraud-detection","url":"https://stripe.com/interviews/fraud-detection","defaultIndex":0,"versions":[{"label":"v1","content":"https://claude.ai/share/55d2d760-85f6-4d0d-87b8-82c2be355627","contentHtml":"

https://claude.ai/share/55d2d760-85f6-4d0d-87b8-82c2be355627

\n","code":"$42","codeHtml":"$43"}],"tags":["type:practical","topic:sliding-window","topic:data-parsing","topic:fraud-detection","difficulty:medium","prep:interview","company:stripe","mode:implement"],"category":"practical","createdAt":1769274368128,"updatedAt":1769274368128,"score":"$undefined","review":"$undefined"},{"id":"499f3e2f-6d96-4570-926d-7ab5bcc1e251","userId":"$undefined","title":"Stripe Practice: Ledger System Design","slug":"stripe-ledger-system","url":"https://stripe.com/interviews/ledger-system","defaultIndex":0,"versions":[{"label":"v1","content":"https://gemini.google.com/app/87c9b7d19320f060","contentHtml":"

https://gemini.google.com/app/87c9b7d19320f060

\n","code":"$44","codeHtml":"$45"}],"tags":["type:practical","topic:idempotency","topic:ledger","topic:sorting","difficulty:medium","prep:interview","company:stripe","mode:design"],"category":"practical","createdAt":1769120785626,"updatedAt":1769120785626,"score":"$undefined","review":"$undefined"},{"id":"16cb9e79-009c-4ddd-a96b-465291fafc2b","userId":"$undefined","title":"1472. Design Browser History","slug":"design-browser-history","url":"https://leetcode.com/problems/design-browser-history/","defaultIndex":0,"versions":[{"label":"v1","content":"Tests our understanding of linear data structures and how to manage \"state\" pointers.\n\nWhile we can solve this using two stacks (one for \"back\" and one for \"forward\"), the most canonical and efficient Python solution uses a dynamic array (List) and a single integer pointer. This approach is generally preferred because it offers O(1) access to any point in the history without moving elements between stacks.\n\n\nhttps://gemini.google.com/app/150083ff0eaf7453","contentHtml":"

Tests our understanding of linear data structures and how to manage \"state\" pointers.

While we can solve this using two stacks (one for \"back\" and one for \"forward\"), the most canonical and efficient Python solution uses a dynamic array (List) and a single integer pointer. This approach is generally preferred because it offers O(1) access to any point in the history without moving elements between stacks.

https://gemini.google.com/app/150083ff0eaf7453

\n","code":"$46","codeHtml":"$47"}],"tags":["type:leetcode","topic:array","topic:linked-list","topic:stack","topic:design","topic:data-stream","difficulty:medium"],"category":"leetcode","createdAt":1767765955039,"updatedAt":1767765955039,"score":"$undefined","review":"$undefined"},{"id":"985634bb-5ccc-42b0-bccb-deb0b0187ff7","userId":"$undefined","title":"2089. Find Target Indices After Sorting Array","slug":"find-target-indices-after-sorting-array","url":"https://leetcode.com/problems/find-target-indices-after-sorting-array/","defaultIndex":0,"versions":[{"label":"v1","content":"While the most direct approach is to sort the array (O(n log n)) and then iterate to find the indices, we can actually solve this in linear time (O(n)) without sorting the entire array.\n\nThe O(n) Counting Strategy:\n\nCount smaller elements: Count how many numbers in the array are strictly less than the target. This count tells us the starting index of the target in a sorted version of the array.\n\nCount occurrences: Count how many times the target itself appears in the array.\n\nGenerate result: Starting from the \"smaller elements\" count, append consecutive integers to our result list for as many times as the target occurred.","contentHtml":"

While the most direct approach is to sort the array (O(n log n)) and then iterate to find the indices, we can actually solve this in linear time (O(n)) without sorting the entire array.

The O(n) Counting Strategy:

Count smaller elements: Count how many numbers in the array are strictly less than the target. This count tells us the starting index of the target in a sorted version of the array.

Count occurrences: Count how many times the target itself appears in the array.

Generate result: Starting from the \"smaller elements\" count, append consecutive integers to our result list for as many times as the target occurred.

\n","code":"class Solution:\n def targetIndices(self, nums: list[int], target: int) -> list[int]:\n count_smaller = 0\n count_target = 0\n \n for num in nums:\n if num < target:\n count_smaller += 1\n elif num == target:\n count_target += 1\n \n # The first index of target will be count_smaller.\n # We generate indices starting from count_smaller for count_target times.\n return [i for i in range(count_smaller, count_smaller + count_target)]","codeHtml":"$48"}],"tags":["type:leetcode","topic:array","topic:binary-search","topic:sorting","difficulty:easy"],"category":"leetcode","createdAt":1767765419284,"updatedAt":1767765419284,"score":"$undefined","review":"$undefined"},{"id":"6907e372-d415-4650-ba46-e7a66717ea5d","userId":"$undefined","title":"974. Subarray Sums Divisible by K","slug":"subarray-sums-divisible-by-k","url":"https://leetcode.com/problems/subarray-sums-divisible-by-k/","defaultIndex":0,"versions":[{"label":"v1","content":"This problem is very similar to **Subarray Sum Equals K**, but instead of looking for a specific sum, we are looking for remainders. If the prefix sum up to index i and index j have the same remainder when divided by K, then the subarray between them must be divisible by K.\n\n### The Logic:\n\n1. **Prefix Sum with Modulo**: As we iterate through the array, maintain a running `prefix_sum`. Calculate the remainder: `remainder = prefix_sum % K`.\n2. **Handle Negative Remainders**: In many programming languages, `-1 % 5` results in `-1`. To ensure we map this correctly, use `(prefix_sum % K + K) % K` to always get a positive remainder.\n3. **Hash Map (Frequency Table)**: Use a map (or an array of size K) to store how many times each remainder has occurred.\n4. **Counting**: If a remainder has appeared times before, it means there are n new subarrays ending at the current index that are divisible by K.","contentHtml":"$49","code":"$4a","codeHtml":"$4b"}],"tags":["type:leetcode","topic:array","topic:hash-table","topic:prefix-sum","difficulty:medium"],"category":"leetcode","createdAt":1767765328723,"updatedAt":1767798919470,"score":"$undefined","review":"$undefined"},{"id":"46a87638-7400-4a78-ab6d-aabc2468459e","userId":"$undefined","title":"1752. Check if Array Is Sorted and Rotated","slug":"check-if-array-is-sorted-and-rotated","url":"https://leetcode.com/problems/check-if-array-is-sorted-and-rotated/","defaultIndex":0,"versions":[{"label":"v1","content":"Allow one","contentHtml":"

Allow one

\n","code":"class Solution:\n def check(self, nums):\n n = len(nums)\n if n <= 2:\n return True\n\n drops = 0\n for i in range(n):\n if nums[i] > nums[(i + 1) % n]: # Wrap-around comparison\n drops += 1\n if drops > 1:\n return False\n return True","codeHtml":"$4c"}],"tags":["type:leetcode","topic:array","difficulty:easy"],"category":"leetcode","createdAt":1767673205651,"updatedAt":1767673205651,"score":"$undefined","review":"$undefined"},{"id":"2741f9c4-bf8b-4cef-98f5-9e69aae3a99e","userId":"$undefined","title":"83. Remove Duplicates from Sorted List","slug":"remove-duplicates-from-sorted-list","url":"https://leetcode.com/problems/remove-duplicates-from-sorted-list/","defaultIndex":0,"versions":[{"label":"v1","content":"Just... it's straightforward.","contentHtml":"

Just... it's straightforward.

\n","code":"# class ListNode:\n# def __init__(self, val=0, next=None):\n# self.val = val\n# self.next = next\nclass Solution:\n def deleteDuplicates(self, head):\n current = head\n\n while current and current.next:\n if current.val == current.next.val:\n # Skip the duplicate node\n current.next = current.next.next\n else:\n # Move forward only when values differ\n current = current.next\n\n return head","codeHtml":"$4d"}],"tags":["type:leetcode","topic:linked-list","difficulty:easy"],"category":"leetcode","createdAt":1767672661632,"updatedAt":1767672661632,"score":"$undefined","review":"$undefined"},{"id":"df34ca5e-12cf-4e82-a168-3cca13d312d2","userId":"$undefined","title":"1838. Frequency of the Most Frequent Element","slug":"frequency-of-the-most-frequent-element","url":"https://leetcode.com/problems/frequency-of-the-most-frequent-element/","defaultIndex":0,"versions":[{"label":"v1","content":"It's about sliding windows.","contentHtml":"

It's about sliding windows.

\n","code":"class Solution:\n def maxFrequency(self, nums, k):\n nums.sort()\n\n left = 0\n window_sum = 0\n best = 1\n\n for right, target in enumerate(nums):\n window_sum += target\n\n # cost to raise nums[left..right] up to nums[right] (target)\n # = target * window_size - window_sum\n while target * (right - left + 1) - window_sum > k:\n window_sum -= nums[left]\n left += 1\n\n best = max(best, right - left + 1)\n\n return best","codeHtml":"$4e"}],"tags":["type:leetcode","topic:array","topic:binary-search","topic:greedy","topic:sliding-window","topic:sorting","topic:prefix-sum","difficulty:medium"],"category":"leetcode","createdAt":1767671602858,"updatedAt":1767671602858,"score":"$undefined","review":"$undefined"},{"id":"c7e6498b-bcc7-4232-af79-3026dae00460","userId":"$undefined","title":"278. First Bad Version","slug":"first-bad-version","url":"https://leetcode.com/problems/first-bad-version/","defaultIndex":0,"versions":[{"label":"v1","content":"This is binary search, but on a monotone predicate instead of searching for a target value.\n\nDefine a boolean function:\n\nP(v) = isBadVersion(v)\n\nGiven the problem statement, P(v) looks like:\n\nFalse, False, False, True, True, True ...\n\nSo we want the first index where P(v) becomes True.\n\nThat corresponds to “bisect_left over a boolean array where False < True”:\n\nbisect_left([F,F,F,T,T], True) -> index of first True","contentHtml":"

This is binary search, but on a monotone predicate instead of searching for a target value.

Define a boolean function:

P(v) = isBadVersion(v)

Given the problem statement, P(v) looks like:

False, False, False, True, True, True ...

So we want the first index where P(v) becomes True.

That corresponds to “bisect_left over a boolean array where False < True”:

bisect_left([F,F,F,T,T], True) -> index of first True

\n","code":"class Solution:\n def firstBadVersion(self, n: int) -> int:\n left, right = 1, n # invariant: answer is in [left, right]\n while left < right:\n mid = left + (right - left) // 2\n if isBadVersion(mid):\n # mid is bad, so the first bad is at mid or to the left\n right = mid\n else:\n # mid is good, so the first bad must be to the right\n left = mid + 1\n return left","codeHtml":"$4f"}],"tags":["type:leetcode","topic:binary-search","topic:interactive","difficulty:easy"],"category":"leetcode","createdAt":1767594197983,"updatedAt":1767594197983,"score":"$undefined","review":"$undefined"},{"id":"883b6030-fdb0-4839-bb6b-adbfe0757c1d","userId":"$undefined","title":"1262. Greatest Sum Divisible by Three","slug":"greatest-sum-divisible-by-three","url":"https://leetcode.com/problems/greatest-sum-divisible-by-three/","defaultIndex":0,"versions":[{"label":"v1","content":"https://gemini.google.com/app/fff1dff1a73639cf","contentHtml":"

https://gemini.google.com/app/fff1dff1a73639cf

\n","code":"class Solution:\n def maxSumDivThree(self, nums):\n # dp[i] = max sum such that sum % 3 == i\n # Start with 0 at index 0, and float('-inf') for others\n dp = [0, float('-inf'), float('-inf')]\n \n for num in nums:\n # Use a temporary list to store updates for this current number\n # so we don't use the same number twice in one transition\n next_dp = dp[:]\n \n for current_sum in dp:\n if current_sum == float('-inf'):\n continue\n \n new_sum = current_sum + num\n remainder = new_sum % 3\n \n # Update the state if the new sum is larger than what we have\n if new_sum > next_dp[remainder]:\n next_dp[remainder] = new_sum\n \n dp = next_dp\n \n return dp[0]","codeHtml":"$50"},{"label":"v2","content":"","contentHtml":"","code":"def maxSumDivThree_Greedy(nums):\n total_sum = sum(nums)\n remainder = total_sum % 3\n \n if remainder == 0:\n return total_sum\n \n # Store the two smallest numbers for remainder 1 and remainder 2\n # Initialized to infinity\n rem1 = sorted([n for n in nums if n % 3 == 1])[:2]\n rem2 = sorted([n for n in nums if n % 3 == 2])[:2]\n \n ans = 0\n if remainder == 1:\n # Option A: Remove the smallest rem1\n # Option B: Remove the two smallest rem2\n res1 = total_sum - rem1[0] if len(rem1) >= 1 else 0\n res2 = total_sum - sum(rem2) if len(rem2) >= 2 else 0\n ans = max(res1, res2)\n \n elif remainder == 2:\n # Option A: Remove the smallest rem2\n # Option B: Remove the two smallest rem1\n res1 = total_sum - rem2[0] if len(rem2) >= 1 else 0\n res2 = total_sum - sum(rem1) if len(rem1) >= 2 else 0\n ans = max(res1, res2)\n \n return ans","codeHtml":"$51"}],"tags":["type:leetcode","topic:array","topic:dynamic-programming","topic:greedy","difficulty:medium"],"category":"leetcode","createdAt":1767593820748,"updatedAt":1767593820748,"score":"$undefined","review":"$undefined"},{"id":"4bc838fd-2c9b-4551-a650-974fa8ddc729","userId":"$undefined","title":"373. Find K Pairs with Smallest Sums","slug":"find-k-pairs-with-smallest-sums","url":"https://leetcode.com/problems/find-k-pairs-with-smallest-sums/","defaultIndex":0,"versions":[{"label":"v1","content":"$52","contentHtml":"$53","code":"class Solution:\n import heapq\n def kSmallestPairs(self, nums1, nums2, k):\n if not nums1 or not nums2 or k <= 0:\n return []\n \n min_heap = []\n res = []\n \n # 1. Initialize the heap with the first element of nums1 \n # paired with the first few elements of nums2.\n # We only need to go up to min(len(nums1), k)\n for i in range(min(len(nums1), k)):\n heapq.heappush(min_heap, (nums1[i] + nums2[0], i, 0))\n \n # 2. Extract the smallest sum and push the next potential candidate\n while min_heap and len(res) < k:\n current_sum, i, j = heapq.heappop(min_heap)\n res.append([nums1[i], nums2[j]])\n \n # If there is a next element in nums2, pair it with the current nums1[i]\n if j + 1 < len(nums2):\n heapq.heappush(min_heap, (nums1[i] + nums2[j+1], i, j+1))\n \n return res","codeHtml":"$54"}],"tags":["type:leetcode","topic:array","topic:heap-priority-queue","difficulty:medium"],"category":"leetcode","createdAt":1767588983408,"updatedAt":1767589346842,"score":"$undefined","review":"$undefined"},{"id":"4c3351ea-a2a1-477a-a52b-df2fcb8ee483","userId":"$undefined","title":"444. Sequence Reconstruction","slug":"sequence-reconstruction","url":"https://leetcode.com/problems/sequence-reconstruction/","defaultIndex":0,"versions":[{"label":"v1","content":"It's about satisfying uniqueness constraints, aka there is only one choice at every given node's prerequisite satisfaction.\n\nWe can think about this in terms of determinism and degrees of freedom. Locks and queues.\n\nThe reward for thinking through these topics isn't in the salary. You are seeking the removal of cognitive load.\n\nThis is what it means to be able and willing to sit with a problem for a long time and think through its implications. No, the solving part isn't important. Understanding is the important part. Making connections is the important part.\n\nhttps://gemini.google.com/app/d0eefaae111c93cd\n\nhttps://chatgpt.com/g/g-68b11a82d24c81918000aa3a81424e5d-coach/c/6959104f-7994-8325-a0d5-0606b36a4555","contentHtml":"

It's about satisfying uniqueness constraints, aka there is only one choice at every given node's prerequisite satisfaction.

We can think about this in terms of determinism and degrees of freedom. Locks and queues.

The reward for thinking through these topics isn't in the salary. You are seeking the removal of cognitive load.

This is what it means to be able and willing to sit with a problem for a long time and think through its implications. No, the solving part isn't important. Understanding is the important part. Making connections is the important part.

https://gemini.google.com/app/d0eefaae111c93cd

https://chatgpt.com/g/g-68b11a82d24c81918000aa3a81424e5d-coach/c/6959104f-7994-8325-a0d5-0606b36a4555

\n","code":"$55","codeHtml":"$56"}],"tags":["type:leetcode","topic:array","topic:graph","topic:topological-sort","difficulty:medium"],"category":"leetcode","createdAt":1767586449343,"updatedAt":1767587948623,"score":"$undefined","review":"$undefined"},{"id":"4a2d16da-9338-47d1-97ba-c5b974d10dba","userId":"$undefined","title":"1376. Time Needed to Inform All Employees","slug":"time-needed-to-inform-all-employees","url":"https://leetcode.com/problems/time-needed-to-inform-all-employees/","defaultIndex":0,"versions":[{"label":"v1","content":"Adjacency lists are used because we're trying to represent a sparse graph. Data and goals and constraints inform structure.\n\nhttps://gemini.google.com/app/8916e0218e61ff66\n\nhttps://chatgpt.com/g/g-68b11a82d24c81918000aa3a81424e5d-coach/c/6958041d-8bf4-832a-b91a-f6b2f718ef35","contentHtml":"

Adjacency lists are used because we're trying to represent a sparse graph. Data and goals and constraints inform structure.

https://gemini.google.com/app/8916e0218e61ff66

https://chatgpt.com/g/g-68b11a82d24c81918000aa3a81424e5d-coach/c/6958041d-8bf4-832a-b91a-f6b2f718ef35

\n","code":"class Solution:\n from collections import deque\n\n def numOfMinutes(self, n, headID, manager, informTime):\n # 1. Build Adjacency List (Manager -> subordinates)\n adj = [[] for _ in range(n)]\n for employee_id, mgr_id in enumerate(manager):\n if mgr_id != -1:\n adj[mgr_id].append(employee_id)\n \n # 2. BFS: (current_employee, cumulative_time_to_reach_them)\n queue = deque([(headID, 0)])\n max_time = 0\n \n while queue:\n curr_emp, curr_time = queue.popleft()\n \n # Update the total time needed\n max_time = max(max_time, curr_time)\n \n # For every subordinate, the time to reach them is \n # current time + the time this manager takes to inform everyone\n for subordinate in adj[curr_emp]:\n queue.append((subordinate, curr_time + informTime[curr_emp]))\n \n return max_time","codeHtml":"$57"}],"tags":["type:leetcode","topic:tree","topic:depth-first-search","topic:breadth-first-search","topic:graph","difficulty:medium"],"category":"leetcode","createdAt":1767382862344,"updatedAt":1767382862344,"score":"$undefined","review":"$undefined"},{"id":"4c932fd5-f55c-46dc-a72e-e57e59f3ce20","userId":"$undefined","title":"2563. Count the Number of Fair Pairs","slug":"count-the-number-of-fair-pairs","url":"https://leetcode.com/problems/count-the-number-of-fair-pairs/","defaultIndex":0,"versions":[{"label":"v1","content":"This problem exemplifies the difference between \"searching for a value\" and \"counting a range.\"\n\nThe indices are a proxy for the count.\n\nSorting turns “which pairs are valid?” into “where is the cutoff index?”, and cutoff indices are countable without enumeration.\n\nWe're looking \"above\" the problem.\n\nThis is the inclusion-exclusion principle, or the prefix sum property.\n\nInstead of trying to stay \"inside\" the narrow window of $[\\text{lower}, \\text{upper}]$, which is hard to manage with two pointers, we solve the \"easier\" problem: \"How many pairs are less than X?\"\n\nTransform a hard condition (“sum in a range”) into something with a monotone boundary (“sum ≤ X”).\n\nThen exploit that boundary to count in bulk.\n\nFinally convert range back via subtraction.\n\nhttps://gemini.google.com/app/fa4236fb21f99f13\n\nhttps://chatgpt.com/g/g-68b11a82d24c81918000aa3a81424e5d-coach/c/6957fce7-f670-8326-b624-c02d219440f2","contentHtml":"$58","code":"class Solution:\n def countFairPairs(self, nums, lower, upper):\n nums.sort()\n \n def countLessEqual(target):\n count = 0\n left = 0\n right = len(nums) - 1\n while left < right:\n if nums[left] + nums[right] <= target:\n # All pairs between left and right work\n count += (right - left)\n left += 1\n else:\n right -= 1\n return count\n \n return countLessEqual(upper) - countLessEqual(lower - 1)","codeHtml":"$59"}],"tags":["type:leetcode","topic:array","topic:two-pointers","topic:binary-search","topic:sorting","difficulty:medium"],"category":"leetcode","createdAt":1767375531366,"updatedAt":1767375531366,"score":"$undefined","review":"$undefined"},{"id":"a0bffe9a-c6ff-49dd-b3a0-8d39147de9ed","userId":"$undefined","title":"179. Largest Number","slug":"largest-number","url":"https://leetcode.com/problems/largest-number/","defaultIndex":0,"versions":[{"label":"v1","content":"We can notice that this is a prefix comparison problem.\n\nhttps://gemini.google.com/app/2017d18d0cef60db\n\nhttps://chatgpt.com/g/g-68b11a82d24c81918000aa3a81424e5d-coach/c/6957f631-9ac0-832b-ba37-32199996c3ed","contentHtml":"

We can notice that this is a prefix comparison problem.

https://gemini.google.com/app/2017d18d0cef60db

https://chatgpt.com/g/g-68b11a82d24c81918000aa3a81424e5d-coach/c/6957f631-9ac0-832b-ba37-32199996c3ed

\n","code":"class Solution:\n from functools import cmp_to_key\n def largestNumber(self, nums):\n # Convert numbers to strings\n num_strs = [str(num) for num in nums]\n \n # Custom comparator\n def compare(a, b):\n if a + b > b + a:\n return -1 # a should come first\n else:\n return 1 # b should come first\n \n # Sort with the custom rule\n num_strs.sort(key=cmp_to_key(compare))\n \n # Join strings together\n result = \"\".join(num_strs)\n \n # Handle the \"000\" edge case by converting to int and back to str\n # Or simply: return '0' if result[0] == '0' else result\n # if parts and parts[0] == \"0\":\n # return \"0\"\n return str(int(result))","codeHtml":"$5a"}],"tags":["type:leetcode","topic:array","topic:string","topic:greedy","topic:sorting","difficulty:medium"],"category":"leetcode","createdAt":1767372821790,"updatedAt":1767372821790,"score":"$undefined","review":"$undefined"},{"id":"0a83f789-8fef-49e3-a908-34c8c0ffa7a7","userId":"$undefined","title":"636. Exclusive Time of Functions","slug":"exclusive-time-of-functions","url":"https://leetcode.com/problems/exclusive-time-of-functions/","defaultIndex":0,"versions":[{"label":"v1","content":"It’s an event-driven simulation. Instead of \"ticking\" a clock second by second (which would be super slow if a function ran for a billion seconds), we only care about the moments when the status quo changes.\n\nhttps://gemini.google.com/app/a509d10809e8ad15\n\nhttps://chatgpt.com/g/g-68b11a82d24c81918000aa3a81424e5d-coach/c/695761f7-f4c4-832e-a750-4112344474e4","contentHtml":"

It’s an event-driven simulation. Instead of \"ticking\" a clock second by second (which would be super slow if a function ran for a billion seconds), we only care about the moments when the status quo changes.

https://gemini.google.com/app/a509d10809e8ad15

https://chatgpt.com/g/g-68b11a82d24c81918000aa3a81424e5d-coach/c/695761f7-f4c4-832e-a750-4112344474e4

\n","code":"$5b","codeHtml":"$5c"}],"tags":["type:leetcode","topic:array","topic:stack","difficulty:medium"],"category":"leetcode","createdAt":1767335687248,"updatedAt":1767335687248,"score":"$undefined","review":"$undefined"},{"id":"6e755d9e-2e10-496e-9c61-8d08bc29effa","userId":"$undefined","title":"841. Keys and Rooms","slug":"keys-and-rooms","url":"https://leetcode.com/problems/keys-and-rooms/","defaultIndex":0,"versions":[{"label":"v1","content":"$5d","contentHtml":"$5e","code":"class Solution:\n def canVisitAllRooms(self, rooms):\n visited = {0}\n stack = [0]\n \n while stack:\n current_room = stack.pop()\n for key in rooms[current_room]:\n if key not in visited:\n visited.add(key)\n stack.append(key)\n \n return len(visited) == len(rooms)","codeHtml":"$5f"}],"tags":["type:leetcode","topic:depth-first-search","topic:breadth-first-search","topic:graph","difficulty:medium"],"category":"leetcode","createdAt":1767333979711,"updatedAt":1767333990148,"score":"$undefined","review":"$undefined"},{"id":"73a31240-40b1-4159-b6e0-4a13f12c026e","userId":"$undefined","title":"525. Contiguous Array","slug":"contiguous-array","url":"https://leetcode.com/problems/contiguous-array/","defaultIndex":0,"versions":[{"label":"v1","content":"There are a lot of interesting connected ideas here on prefixes and monoticity and \"imposing\" a bit of monoticity onto a problem which actually doesn't ask for it.\n\nIt feels like a sliding window, but sliding windows only work on window properties that are monotone with respect to expanding the window.\n\nThe unifying move behind a huge chunk of \"array algorithm\" problems is:\n\n> Turn a messy global problem into local bookkeeping where cancellation, balance, or repetition reveals structure.\n\nThere are a lot of invariants. That's the thing that we are exploiting. That's why they all feel similar.\n\nhttps://gemini.google.com/app/0c0ecf98e21e9e10\n\nhttps://chatgpt.com/g/g-68b11a82d24c81918000aa3a81424e5d-coach/c/6950c926-0324-832d-a367-0d0e300783c8","contentHtml":"

There are a lot of interesting connected ideas here on prefixes and monoticity and \"imposing\" a bit of monoticity onto a problem which actually doesn't ask for it.

It feels like a sliding window, but sliding windows only work on window properties that are monotone with respect to expanding the window.

The unifying move behind a huge chunk of \"array algorithm\" problems is:

\n
Turn a messy global problem into local bookkeeping where cancellation, balance, or repetition reveals structure.
\n

There are a lot of invariants. That's the thing that we are exploiting. That's why they all feel similar.

https://gemini.google.com/app/0c0ecf98e21e9e10

https://chatgpt.com/g/g-68b11a82d24c81918000aa3a81424e5d-coach/c/6950c926-0324-832d-a367-0d0e300783c8

\n","code":"class Solution:\n def findMaxLength(self, nums):\n # Map to store (prefix_sum: first_index_it_occurred)\n # Initialize with 0: -1 to handle subarrays starting from index 0\n indices = {0: -1}\n max_len = 0\n count = 0\n \n for i, num in enumerate(nums):\n # Treat 0 as -1 and 1 as 1\n count += 1 if num == 1 else -1\n \n if count in indices:\n # Calculate length: current index - first occurrence of this sum\n max_len = max(max_len, i - indices[count])\n else:\n # Only store the first occurrence to maximize the distance/length\n indices[count] = i\n \n return max_len","codeHtml":"$60"}],"tags":["type:leetcode","topic:array","topic:hash-table","topic:prefix-sum","difficulty:medium"],"category":"leetcode","createdAt":1766902875919,"updatedAt":1766903054328,"score":"$undefined","review":"$undefined"},{"id":"375d1705-631c-4c22-bdb2-3b4d53fe529b","userId":"$undefined","title":"29. Divide Two Integers","slug":"divide-two-integers","url":"https://leetcode.com/problems/divide-two-integers/","defaultIndex":0,"versions":[{"label":"v1","content":"Bitwise division. If it's divisible by k, maybe it's divisible by 2k, then 4k, and so on.\n\nhttps://gemini.google.com/app/74b2951b2feecf51\n\nhttps://chatgpt.com/g/g-68b11a82d24c81918000aa3a81424e5d-coach/c/6950c623-2a44-8330-a098-cd0ed4f6bb28","contentHtml":"

Bitwise division. If it's divisible by k, maybe it's divisible by 2k, then 4k, and so on.

https://gemini.google.com/app/74b2951b2feecf51

https://chatgpt.com/g/g-68b11a82d24c81918000aa3a81424e5d-coach/c/6950c623-2a44-8330-a098-cd0ed4f6bb28

\n","code":"class Solution:\n def divide(self, dividend: int, divisor: int) -> int:\n # Constants for 32-bit range\n MAX_INT = 2147483647\n MIN_INT = -2147483648\n\n # Edge case: Overflow\n if dividend == MIN_INT and divisor == -1:\n return MAX_INT\n\n # Determine sign\n negative = (dividend < 0) ^ (divisor < 0)\n \n # Use absolute values\n dividend, divisor = abs(dividend), abs(divisor)\n quotient = 0\n\n # Exponential subtraction\n while dividend >= divisor:\n temp_divisor = divisor\n multiple = 1\n # Double the divisor as much as possible\n while dividend >= (temp_divisor << 1):\n temp_divisor <<= 1\n multiple <<= 1\n \n # Subtract the largest found chunk\n dividend -= temp_divisor\n quotient += multiple\n\n return -quotient if negative else quotient","codeHtml":"$61"}],"tags":["type:leetcode","topic:math","topic:bit-manipulation","difficulty:medium"],"category":"leetcode","createdAt":1766901848426,"updatedAt":1766901848426,"score":"$undefined","review":"$undefined"},{"id":"e744e276-8cf7-4e9a-a176-50aed71ae519","userId":"$undefined","title":"1015. Smallest Integer Divisible by K","slug":"smallest-integer-divisible-by-k","url":"https://leetcode.com/problems/smallest-integer-divisible-by-k/","defaultIndex":0,"versions":[{"label":"v1","content":"$62","contentHtml":"$63","code":"class Solution:\n def smallestRepunitDivByK(self, k: int) -> int:\n # If k is even or divisible by 5, no number ending in 1 can be divisible by it\n if k % 2 == 0 or k % 5 == 0:\n return -1\n \n remainder = 0\n # The length of n can at most be k\n for length in range(1, k + 1):\n # Update the remainder: (previous_remainder * 10 + 1) % k\n remainder = (remainder * 10 + 1) % k\n \n # If remainder is 0, we found the smallest n\n if remainder == 0:\n return length\n \n return -1","codeHtml":"$64"}],"tags":["type:leetcode","topic:hash-table","topic:math","difficulty:medium"],"category":"leetcode","createdAt":1766901019723,"updatedAt":1766901019723,"score":"$undefined","review":"$undefined"},{"id":"336aac14-16a1-4bd2-9b6a-973f3c5e8f9e","userId":"$undefined","title":"34. Find First and Last Position of Element in Sorted Array","slug":"find-first-and-last-position-of-element-in-sorted-array","url":"https://leetcode.com/problems/find-first-and-last-position-of-element-in-sorted-array/","defaultIndex":0,"versions":[{"label":"v1","content":"Non-decreasing is actually code for \"sorted, but there might be duplicates.\"\n\nhttps://gemini.google.com/app/f8bd4a7938802a78","contentHtml":"

Non-decreasing is actually code for \"sorted, but there might be duplicates.\"

https://gemini.google.com/app/f8bd4a7938802a78

\n","code":"class Solution:\n def searchRange(self, nums, target):\n def find_first(nums, target):\n low, high = 0, len(nums) - 1\n first_idx = -1\n while low <= high:\n mid = (low + high) // 2\n if nums[mid] >= target:\n if nums[mid] == target:\n first_idx = mid\n high = mid - 1\n else:\n low = mid + 1\n return first_idx\n\n def find_last(nums, target):\n low, high = 0, len(nums) - 1\n last_idx = -1\n while low <= high:\n mid = (low + high) // 2\n if nums[mid] <= target:\n if nums[mid] == target:\n last_idx = mid\n low = mid + 1\n else:\n high = mid - 1\n return last_idx\n\n return [find_first(nums, target), find_last(nums, target)]","codeHtml":"$65"},{"label":"v2","content":"bisect_left gives you the first index where the target could be placed while maintaining order.\n\nbisect_right gives you the last index + 1.\n\nThe only catch is that bisect doesn't check if the target actually exists in the list; it just tells us where it would go. We'd have to add a manual check.\n\nEdge Cases: If the list is empty or the target isn't present, the manual search naturally returns -1. In the bisect version, we must verify the index is within range(len(nums)).","contentHtml":"

bisect_left gives you the first index where the target could be placed while maintaining order.

bisect_right gives you the last index + 1.

The only catch is that bisect doesn't check if the target actually exists in the list; it just tells us where it would go. We'd have to add a manual check.

Edge Cases: If the list is empty or the target isn't present, the manual search naturally returns -1. In the bisect version, we must verify the index is within range(len(nums)).

\n","code":"class Solution:\n def searchRange(self, nums, target):\n left = bisect.bisect_left(nums, target)\n right = bisect.bisect_right(nums, target) - 1\n \n # Check if left is within bounds and actually points to the target\n if left < len(nums) and nums[left] == target:\n return [left, right]\n \n return [-1, -1]","codeHtml":"$66"}],"tags":["type:leetcode","topic:array","topic:binary-search","difficulty:medium"],"category":"leetcode","createdAt":1766900072410,"updatedAt":1767765045614,"score":"$undefined","review":"$undefined"},{"id":"5f727390-5235-49c5-b6ea-2de3b4bb9856","userId":"$undefined","title":"327. Count of Range Sum","slug":"count-of-range-sum","url":"https://leetcode.com/problems/count-of-range-sum/","defaultIndex":0,"versions":[{"label":"v1","content":"I'm just putting this here for posterity. We should expect to not see this, but there're a lot of thoughts I've been having about the essence of programming, and what it means to seek truth.\n\nI've found that two axes can relatively nicely distil what I've been thinking:\n\n1. We want to reduce ambiguity — this means exhaustively mapping out the cases. Following the implicit structure and making it explicit, or imposing a known algorithm or structure to make a problem conform to a known solution.\n\n2. We want to reduce complexity — this comes in the form of identifying places where we can make use of simplicity. What's true is often more simple, relatively speaking.\n\nhttps://gemini.google.com/app/996e78a63b751252\n\nhttps://chatgpt.com/g/g-68b11a82d24c81918000aa3a81424e5d-coach/c/694c8197-65c8-8328-92b7-5f4cc6378ba5","contentHtml":"$67","code":"$68","codeHtml":"$69"}],"tags":["type:leetcode","topic:array","topic:binary-search","topic:divide-and-conquer","topic:binary-indexed-tree","topic:segment-tree","topic:merge-sort","topic:ordered-set","difficulty:hard"],"category":"leetcode","createdAt":1766625041059,"updatedAt":1766897959637,"score":"$undefined","review":"$undefined"},{"id":"883afabb-4da4-4fd9-af53-a4d6a3f96191","userId":"$undefined","title":"493. Reverse Pairs","slug":"reverse-pairs","url":"https://leetcode.com/problems/reverse-pairs/","defaultIndex":0,"versions":[{"label":"v1","content":"It’s like a tournament bracket:\n\n1. Divide the players into two brackets.\n\n2. Sort the players within their own brackets (Recursion).\n\n3. Compare players from Bracket A against Bracket B to find reverse pairs (The Counting Step). We can do this fast because they are sorted!\n\n4. Merge the brackets into one big ranked list (The Sorting Step) so the next level up can repeat the process.\n\nThis is pretty nuanced. There's a logical leap here, and though the code's simple, the divide and conquer strategy warrants close investigation.\n\nSorting preserves relative order between two halves because... we're sorting the two halves. Not the whole thing.\n\n\nhttps://gemini.google.com/app/4e48c919d68f3326","contentHtml":"

It’s like a tournament bracket:

Divide the players into two brackets.
\n
Sort the players within their own brackets (Recursion).
\n
Compare players from Bracket A against Bracket B to find reverse pairs (The Counting Step). We can do this fast because they are sorted!
\n
Merge the brackets into one big ranked list (The Sorting Step) so the next level up can repeat the process.
\n

This is pretty nuanced. There's a logical leap here, and though the code's simple, the divide and conquer strategy warrants close investigation.

Sorting preserves relative order between two halves because... we're sorting the two halves. Not the whole thing.

https://gemini.google.com/app/4e48c919d68f3326

\n","code":"class Solution:\n def reversePairs(self, nums: list[int]) -> int:\n def merge_sort(start, end):\n if start >= end:\n return 0\n \n mid = (start + end) // 2\n count = merge_sort(start, mid) + merge_sort(mid + 1, end)\n \n # --- Counting Step ---\n # Use two pointers to count reverse pairs between the two sorted halves\n j = mid + 1\n for i in range(start, mid + 1):\n while j <= end and nums[i] > 2 * nums[j]:\n j += 1\n count += (j - (mid + 1))\n \n # --- Standard Merge Step ---\n nums[start:end+1] = sorted(nums[start:end+1])\n return count\n\n return merge_sort(0, len(nums) - 1)","codeHtml":"$6a"}],"tags":["type:leetcode","topic:array","topic:binary-search","topic:divide-and-conquer","topic:binary-indexed-tree","topic:segment-tree","topic:merge-sort","topic:ordered-set","difficulty:hard"],"category":"leetcode","createdAt":1766537547776,"updatedAt":1766538879229,"score":"$undefined","review":"$undefined"},{"id":"396c00c8-c1bd-43e0-9aef-fded211d8961","userId":"$undefined","title":"1255. Maximum Score Words Formed by Letters","slug":"maximum-score-words-formed-by-letters","url":"https://leetcode.com/problems/maximum-score-words-formed-by-letters/","defaultIndex":0,"versions":[{"label":"v1","content":"This problem is a variation of the Knapsack Problem or the Subset Sum Problem, but with character frequency constraints. Since the number of words is relatively small, we can solve it using Backtracking or Bitmasking to explore all possible subsets of words.\n\nFor each word, we have two choices:\n\nSkip the word: Move to the next word without using any letters.\n\nUse the word: Check if we have enough characters in our letters supply. If we do, subtract them, add the word's score, and recurse. We \"undo\" the subtraction (backtrack) after the recursive call.\n\nhttps://gemini.google.com/app/c35f13e0db2b50b9","contentHtml":"

This problem is a variation of the Knapsack Problem or the Subset Sum Problem, but with character frequency constraints. Since the number of words is relatively small, we can solve it using Backtracking or Bitmasking to explore all possible subsets of words.

For each word, we have two choices:

Skip the word: Move to the next word without using any letters.

Use the word: Check if we have enough characters in our letters supply. If we do, subtract them, add the word's score, and recurse. We \"undo\" the subtraction (backtrack) after the recursive call.

https://gemini.google.com/app/c35f13e0db2b50b9

\n","code":"$6b","codeHtml":"$6c"}],"tags":["type:leetcode","topic:array","topic:string","topic:dynamic-programming","topic:backtracking","topic:bit-manipulation","topic:bitmask","difficulty:hard"],"category":"leetcode","createdAt":1766453393660,"updatedAt":1766898056442,"score":"$undefined","review":"$undefined"},{"id":"838c842f-9e4b-4620-b6f0-4e4b72a0e06d","userId":"$undefined","title":"12. Integer to Roman","slug":"integer-to-roman","url":"https://leetcode.com/problems/integer-to-roman/","defaultIndex":0,"versions":[{"label":"v1","content":"It's intimidating, but it's simply a mapping exercise. 4s and 9s are simply additional symbols in a bucket.\n\nIt's very concatenative. It's very no-brainer. To add, they simply merge them together, sort them, and simplify by the rule of 5.\n\nIt's a very good system for when you're in a low-literacy society.\n\nhttps://gemini.google.com/app/6016cadbff3c38ff","contentHtml":"

It's intimidating, but it's simply a mapping exercise. 4s and 9s are simply additional symbols in a bucket.

It's very concatenative. It's very no-brainer. To add, they simply merge them together, sort them, and simplify by the rule of 5.

It's a very good system for when you're in a low-literacy society.

https://gemini.google.com/app/6016cadbff3c38ff

\n","code":"class Solution:\n def intToRoman(self, num: int) -> str:\n # Define the mapping in descending order\n roman_map = [\n (1000, \"M\"), (900, \"CM\"), (500, \"D\"), (400, \"CD\"),\n (100, \"C\"), (90, \"XC\"), (50, \"L\"), (40, \"XL\"),\n (10, \"X\"), (9, \"IX\"), (5, \"V\"), (4, \"IV\"), (1, \"I\")\n ]\n \n result = []\n \n for value, symbol in roman_map:\n # Check how many times the current symbol fits into num\n count = num // value\n if count > 0:\n result.append(symbol * count)\n # Reduce num by the amount we just converted\n num %= value\n \n return \"\".join(result)","codeHtml":"$6d"}],"tags":["type:leetcode","topic:hash-table","topic:math","topic:string","difficulty:medium"],"category":"leetcode","createdAt":1766452250143,"updatedAt":1766452250143,"score":"$undefined","review":"$undefined"},{"id":"126b964b-f704-4789-a419-b4c062de6b7c","userId":"$undefined","title":"257. Binary Tree Paths","slug":"binary-tree-paths","url":"https://leetcode.com/problems/binary-tree-paths/","defaultIndex":0,"versions":[{"label":"v1","content":"It's all about carrying state (the current path) through a recursion tree and recognizing the terminal condition.\n\nFor this question, top-down (passing the string path down) is quite intuitive.","contentHtml":"

It's all about carrying state (the current path) through a recursion tree and recognizing the terminal condition.

For this question, top-down (passing the string path down) is quite intuitive.

\n","code":"class Solution:\n def binaryTreePaths(self, root):\n def dfs(node, path_list):\n if not node:\n return\n \n # 1. Add current value to the list\n path_list.append(str(node.val))\n \n # 2. Check for leaf\n if not node.left and not node.right:\n res.append(\"->\".join(path_list))\n else:\n # 3. Recurse down\n dfs(node.left, path_list)\n dfs(node.right, path_list)\n \n # 4. BACKTRACK: Remove the node so it doesn't leak into the next path\n path_list.pop()\n\n res = []\n dfs(root, [])\n return res","codeHtml":"$6e"}],"tags":["type:leetcode","topic:string","topic:backtracking","topic:tree","topic:depth-first-search","topic:binary-tree","difficulty:easy"],"category":"leetcode","createdAt":1766450780251,"updatedAt":1766450780251,"score":"$undefined","review":"$undefined"},{"id":"3f9023e2-ed3e-40f2-9700-8ce429e5fa3c","userId":"$undefined","title":"437. Path Sum III","slug":"path-sum-iii","url":"https://leetcode.com/problems/path-sum-iii/","defaultIndex":0,"versions":[{"label":"v1","content":"We have to take a leap of faith with induction. The dfs function promises that when it finishes executing, the prefix_sums map will be in the exact same state it was in before the function started.\n\nWe add one to the map, and we propagate it down to the branches. And we pop it off. And we trust that the children follow the instructions of the parent, because they hold the exact instruction sets. \n\nWe end up with a net 0 prefix_sums map at the end.\n\nOther than the 0: 1 entry. That's the ground truth.\n\nThere's a lot of semantic meaning in the +1 and -1. These aren't merely isolated operations. These are the physical boundaries of the subproblem's scope.\n\nOnce we trust the induction, the code is trivial. We don't have to worry about \"What if the left child changes the map?\" because we know that the post-condition prevents that. We can focus entirely on the current node's responsibility:\n\n1. Check the mile markers above you.\n\n2. Leave your own mile marker for those above you.\n\n3. Clean up before you leave.","contentHtml":"$6f","code":"$70","codeHtml":"$71"}],"tags":["type:leetcode","topic:tree","topic:depth-first-search","topic:binary-tree","difficulty:medium"],"category":"leetcode","createdAt":1766367054660,"updatedAt":1766367054660,"score":"$undefined","review":"$undefined"},{"id":"1d3806ea-576f-499f-9f2c-7d22e1389b9b","userId":"$undefined","title":"112. Path Sum","slug":"path-sum","url":"https://leetcode.com/problems/path-sum/","defaultIndex":0,"versions":[{"label":"v1","content":"This is actually quite elegant as a thing to study. \n\nTree problems are about asking: \"What is the one question I need to answer at each node, and how do I transform the input so that question is trivial to ask?\"\n\nThe kernel of recursion is delegation and local perspective. We reduce complexity as much as possible along each step of the way.","contentHtml":"

This is actually quite elegant as a thing to study.

Tree problems are about asking: \"What is the one question I need to answer at each node, and how do I transform the input so that question is trivial to ask?\"

The kernel of recursion is delegation and local perspective. We reduce complexity as much as possible along each step of the way.

\n","code":"class Solution:\n def hasPathSum(self, root: Optional[TreeNode], targetSum: int) -> bool:\n if not root:\n return False\n \n # If it's a leaf, it has to be the value\n if not root.left and not root.right:\n return root.val == targetSum\n \n remaining = targetSum - root.val\n return self.hasPathSum(root.left, remaining) or self.hasPathSum(root.right, remaining)","codeHtml":"$72"}],"tags":["type:leetcode","topic:tree","topic:depth-first-search","topic:breadth-first-search","topic:binary-tree","difficulty:easy"],"category":"leetcode","createdAt":1766361501613,"updatedAt":1766361501613,"score":"$undefined","review":"$undefined"},{"id":"981c017c-a175-4d8e-8b92-ef7bae2b2fab","userId":"$undefined","title":"1110. Delete Nodes And Return Forest","slug":"delete-nodes-and-return-forest","url":"https://leetcode.com/problems/delete-nodes-and-return-forest/","defaultIndex":0,"versions":[{"label":"v1","content":"$73","contentHtml":"$74","code":"$75","codeHtml":"$76"}],"tags":["type:leetcode","topic:array","topic:hash-table","topic:tree","topic:depth-first-search","topic:binary-tree","difficulty:medium"],"category":"leetcode","createdAt":1766279968852,"updatedAt":1766279968852,"score":"$undefined","review":"$undefined"},{"id":"af8839d7-9385-442e-ba54-dbf1bdcd122a","userId":"$undefined","title":"852. Peak Index in a Mountain Array","slug":"peak-index-in-a-mountain-array","url":"https://leetcode.com/problems/peak-index-in-a-mountain-array/","defaultIndex":0,"versions":[{"label":"v1","content":"Why this works\nSearch Space: We start with the whole array.\n\nThe Comparison: We compare arr[mid] to its neighbor. This essentially calculates a \"local derivative.\"\n\nThe Convergence: Because the array is guaranteed to be a mountain, the left pointer will climb the hill from the left, and the right pointer will descend the hill from the right until they meet exactly at the highest point.\n\nWe're simply looking at the derivative!\nAnd derivatives are just... they go in both directions!\n\n\"Since this is a unimodal function, I'm going to perform a binary search on the discrete derivative to find the root where the slope flips\"\n\nhttps://gemini.google.com/app/f376467b35ba9b54","contentHtml":"

Why this works
Search Space: We start with the whole array.

The Comparison: We compare arr[mid] to its neighbor. This essentially calculates a \"local derivative.\"

The Convergence: Because the array is guaranteed to be a mountain, the left pointer will climb the hill from the left, and the right pointer will descend the hill from the right until they meet exactly at the highest point.

We're simply looking at the derivative!
And derivatives are just... they go in both directions!

\"Since this is a unimodal function, I'm going to perform a binary search on the discrete derivative to find the root where the slope flips\"

https://gemini.google.com/app/f376467b35ba9b54

\n","code":"class Solution:\n def peakIndexInMountainArray(self, arr: List[int]) -> int:\n left, right = 0, len(arr) - 1\n while left < right:\n mid = (left + right) // 2\n\n # Check if we're on an uphill or a downhill\n if arr[mid] < arr[mid + 1]:\n # We're on the left side of the peak! Move right\n left = mid + 1\n else:\n # We're either at the peak or on the right side! Move left\n # We can't do mid - 1 because mid might be the peak itself\n right = mid\n return left","codeHtml":"$77"}],"tags":["type:leetcode","topic:array","topic:binary-search","difficulty:medium"],"category":"leetcode","createdAt":1766278971030,"updatedAt":1766278971030,"score":"$undefined","review":"$undefined"},{"id":"a6e8962a-4af7-484d-8beb-e9b2e4bb62bb","userId":"$undefined","title":"931. Minimum Falling Path Sum","slug":"minimum-falling-path-sum","url":"https://leetcode.com/problems/minimum-falling-path-sum/","defaultIndex":0,"versions":[{"label":"v1","content":"$78","contentHtml":"$79","code":"class Solution:\n def minFallingPathSum(self, matrix: List[List[int]]) -> int:\n rows = len(matrix)\n cols = len(matrix[0])\n\n # Start from the second row (index 1)\n for r in range(1, rows):\n for c in range(cols):\n # 1) Directly above\n best_above = matrix[r-1][c]\n\n # 2) Diagonal left maybe\n if c > 0:\n best_above = min(best_above, matrix[r-1][c-1])\n \n # 3) Diagonal right maybe\n if c < cols - 1:\n best_above = min(best_above, matrix[r-1][c+1])\n \n # Update current cell\n matrix[r][c] += best_above\n return min(matrix[-1])","codeHtml":"$7a"}],"tags":["type:leetcode","topic:array","topic:dynamic-programming","topic:matrix","difficulty:medium"],"category":"leetcode","createdAt":1765925749476,"updatedAt":1765925749476,"score":"$undefined","review":"$undefined"},{"id":"57fc3117-4626-441f-a4fd-cffe5ae82b4d","userId":"$undefined","title":"154. Find Minimum in Rotated Sorted Array II","slug":"find-minimum-in-rotated-sorted-array-ii","url":"https://leetcode.com/problems/find-minimum-in-rotated-sorted-array-ii/","defaultIndex":0,"versions":[{"label":"v1","content":"We move right and not mid because... we need to shrink the window. That's it, really. It's remarkably similar to the first one. Having duplicates is... a 2-line change, ish.\n\nIt's similar to:\n\nhttps://leetcode.com/problems/search-in-rotated-sorted-array-ii/\n\nSearch in Rotated Sorted Array II\n\nhttps://gemini.google.com/app/06d1537eada50f6a","contentHtml":"

We move right and not mid because... we need to shrink the window. That's it, really. It's remarkably similar to the first one. Having duplicates is... a 2-line change, ish.

Search in Rotated Sorted Array II

https://gemini.google.com/app/06d1537eada50f6a

\n","code":"class Solution:\n def findMin(self, nums: List[int]) -> int:\n left, right = 0, len(nums) - 1\n while left < right:\n mid = (left + right) // 2\n\n if nums[mid] < nums[right]:\n # Minimum is in [left..mid]\n right = mid\n elif nums[mid] > nums[right]:\n # Minimum is in (mid..right]\n left = mid + 1\n else:\n # Equal, so shrink the window\n right -= 1\n return nums[left]","codeHtml":"$7b"}],"tags":["type:leetcode","topic:array","topic:binary-search","difficulty:hard"],"category":"leetcode","createdAt":1765921687063,"updatedAt":1765921823347,"score":"$undefined","review":"$undefined"},{"id":"b45c3e83-5e4a-47f5-b9c3-2355051302f1","userId":"$undefined","title":"1044. Longest Duplicate Substring","slug":"longest-duplicate-substring","url":"https://leetcode.com/problems/longest-duplicate-substring/","defaultIndex":0,"versions":[{"label":"v1","content":"I have no idea what's going on bro. I know Binary Search on Answer, and allegedly, the rolling hash thing is \"Sliding Window Sum on Steroids.\"\n\nWe're just updating a value as we slide a window.","contentHtml":"

I have no idea what's going on bro. I know Binary Search on Answer, and allegedly, the rolling hash thing is \"Sliding Window Sum on Steroids.\"

We're just updating a value as we slide a window.

\n","code":"$7c","codeHtml":"$7d"}],"tags":["type:leetcode","topic:string","topic:binary-search","topic:hash-function","topic:rolling-hash","topic:sliding-window","topic:suffix-array","difficulty:hard"],"category":"leetcode","createdAt":1765844289893,"updatedAt":1765844289893,"score":"$undefined","review":"$undefined"},{"id":"b6a524c5-22f0-449e-b07b-7bce50f19814","userId":"$undefined","title":"402. Remove K Digits","slug":"remove-k-digits","url":"https://leetcode.com/problems/remove-k-digits/","defaultIndex":0,"versions":[{"label":"v1","content":"$7e","contentHtml":"$7f","code":"class Solution:\n def removeKdigits(self, num: str, k: int) -> str:\n stack = []\n\n for digit in num:\n # While we have k to spend, the stack isn't empty,\n # and the previous digit is bigger than the current one:\n while k > 0 and stack and stack[-1] > digit:\n stack.pop()\n k -= 1\n stack.append(digit)\n # Edge Case 2: If the numbers were already ascending (e.g. '12345', k=2),\n # we might still have k left to remove. Remove from the end.\n if k > 0:\n stack = stack[:-k]\n # Join the stack to form the string\n result = ''.join(stack)\n\n # Edge case 3: remove leading zeroes\n result = result.lstrip('0')\n\n # Edge case 4: If the string is empty (we removed everything), return '0'\n return result if result else '0'","codeHtml":"$80"}],"tags":["type:leetcode","topic:stack","topic:greedy","topic:string","topic:monotonic-stack","difficulty:medium"],"category":"leetcode","createdAt":1765843001868,"updatedAt":1765843001868,"score":"$undefined","review":"$undefined"},{"id":"d312a40e-6a4e-4b55-88ab-1b309af0ef0f","userId":"$undefined","title":"123. Best Time to Buy and Sell Stock III","slug":"best-time-to-buy-and-sell-stock-iii","url":"https://leetcode.com/problems/best-time-to-buy-and-sell-stock-iii/","defaultIndex":0,"versions":[{"label":"v1","content":"Since you can make at most two transactions and they cannot overlap, imagine splitting the array of prices into two parts at a specific day i.\n\nThe problem can be split into two sub-problems. If we are allowed two transactions, there must be some dividing line, say day i, where:\n\n1. Transaction 1 happens on or before day i.\n2. Transaction 2 happens on or after day i.\n\nBy iterating through every possible split point i and asking \"What is the max profit on the left + max profit on the right?\", you guarantee finding the global maximum.\n\nTo transition into the O(1) space solution, we can think:\n\nSince Transaction 2 occurs after Transaction 1, we don't need to store the entire history of Transaction 1. We can just calculate the current state of T1 and feed it directly into the cost basis of T2 in the same loop. This removes the need for the array.\n\nhttps://gemini.google.com/app/b5d7100d95f3bfaa\n\nhttps://aistudio.google.com/prompts/1Lzk91exDfjXevJyCuhachpqQvj3eabzy","contentHtml":"$81","code":"$82","codeHtml":"$83"},{"label":"v2","content":"$84","contentHtml":"$85","code":"class Solution:\n def maxProfit(self, prices: List[int]) -> int:\n # Initialize costs to infinity and profits to 0\n buy1 = float('inf') # Cost of first stock\n profit1 = 0 # Profit after selling first stock\n\n buy2 = float('inf') # Cost of second stock (reinvested)\n profit2 = 0 # Profit after selling second stock\n\n for price in prices:\n # 1) Minimize the cost of the first stock\n buy1 = min(buy1, price)\n\n # 2) Maximize profit of the first transaction\n profit1 = max(profit1, price - buy1)\n\n # 3) Minimize cost of second stock\n # (Treat previous profit as a discount on the second buy price)\n buy2 = min(buy2, price - profit1)\n\n # 4) Maximize total profit after second transaction\n profit2 = max(profit2, price - buy2)\n return profit2","codeHtml":"$86"}],"tags":["type:leetcode","topic:array","topic:dynamic-programming","difficulty:hard"],"category":"leetcode","createdAt":1765839005251,"updatedAt":1766898217782,"score":"$undefined","review":"$undefined"},{"id":"6c3b900e-d6e0-4e65-b705-ffdf57ad7126","userId":"$undefined","title":"1326. Minimum Number of Taps to Open to Water a Garden","slug":"minimum-number-of-taps-to-open-to-water-a-garden","url":"https://leetcode.com/problems/minimum-number-of-taps-to-open-to-water-a-garden/","defaultIndex":0,"versions":[{"label":"v1","content":"$87","contentHtml":"$88","code":"$89","codeHtml":"$8a"},{"label":"v2","content":"See Jump Game.","contentHtml":"

See Jump Game.

\n","code":"$8b","codeHtml":"$8c"}],"tags":["type:leetcode","topic:greedy","topic:array","topic:dynamic-programming","difficulty:hard"],"category":"leetcode","createdAt":1765834611965,"updatedAt":1765834611965,"score":"$undefined","review":"$undefined"},{"id":"8aa3dd18-86ab-4b2b-9e2b-30ccfa6641f1","userId":"$undefined","title":"1024. Video Stitching","slug":"video-stitching","url":"https://leetcode.com/problems/video-stitching/","defaultIndex":0,"versions":[{"label":"v1","content":"This is a greedy on intervals, which is a sort of optimized BFS.\n\nThis is like jump game. You can see jump game as bfs because you can look at the farthest reachable places as demarcating the ends of levels.","contentHtml":"

This is a greedy on intervals, which is a sort of optimized BFS.

This is like jump game. You can see jump game as bfs because you can look at the farthest reachable places as demarcating the ends of levels.

\n","code":"class Solution:\n def videoStitching(self, clips: List[List[int]], time: int) -> int:\n # Sort by start time.\n # End tie-breaking doesn't matter if we take max end during the scan\n clips.sort()\n\n count = 0\n current_end = 0\n farthest_reach = 0\n\n i = 0\n n = len(clips)\n\n # We continue as long as we haven't covered the whole time\n while current_end < time:\n # Process all clips that start before or when the current clip ends\n while i < n and clips[i][0] <= current_end:\n farthest_reach = max(farthest_reach, clips[i][1])\n i += 1\n # If we couldn't find a clip to extend our range, we're stuck!\n if farthest_reach == current_end:\n return -1\n # Let's extend to the farthest possible point!\n current_end = farthest_reach\n count += 1\n return count","codeHtml":"$8d"},{"label":"v2","content":"","contentHtml":"","code":"def videoStitching(clips, time):\n clips.sort()\n \n # BFS Variables\n current_level_boundary = 0\n next_level_boundary = 0\n jumps = 0\n \n i = 0\n while current_level_boundary < time:\n # Process the \"Queue\" (all clips start within current range)\n while i < len(clips) and clips[i][0] <= current_level_boundary:\n next_level_boundary = max(next_level_boundary, clips[i][1])\n i += 1\n \n # Dead End Check (Queue empty)\n if current_level_boundary == next_level_boundary:\n return -1\n \n # Move to next level\n current_level_boundary = next_level_boundary\n jumps += 1\n \n return jumps","codeHtml":"$8e"}],"tags":["type:leetcode","topic:array","topic:dynamic-programming","topic:greedy","difficulty:medium"],"category":"leetcode","createdAt":1765608656998,"updatedAt":1765608656998,"score":"$undefined","review":"$undefined"},{"id":"aeaf2799-190b-4d2a-863f-951685608efd","userId":"$undefined","title":"689. Maximum Sum of 3 Non-Overlapping Subarrays","slug":"maximum-sum-of-3-non-overlapping-subarrays","url":"https://leetcode.com/problems/maximum-sum-of-3-non-overlapping-subarrays/","defaultIndex":0,"versions":[{"label":"v1","content":"Pretty straightforward. Precompute the left and right, then combine. ","contentHtml":"

Pretty straightforward. Precompute the left and right, then combine.

\n","code":"$8f","codeHtml":"$90"}],"tags":["type:leetcode","topic:array","topic:dynamic-programming","difficulty:hard"],"category":"leetcode","createdAt":1765603011939,"updatedAt":1765603011939,"score":"$undefined","review":"$undefined"},{"id":"5234f1f0-a93c-4da3-965e-d93790d3a377","userId":"$undefined","title":"1574. Shortest Subarray to be Removed to Make Array Sorted","slug":"shortest-subarray-to-be-removed-to-make-array-sorted","url":"https://leetcode.com/problems/shortest-subarray-to-be-removed-to-make-array-sorted/","defaultIndex":0,"versions":[{"label":"v1","content":"This is actually mostly an index hygiene question. \n\nBounds check first, value check second.\n\nWe have to calculate the \"elements removed\" with a - 1 because we don't want to count the destination.\n\nWe're looking for the open interval. Therefore, we need to take off the destination.\n\nThis... goes pretty deep. I'm not even talking about the algorithm here, which is just a kind of the merge of merge sort, which is just comparing numbers.\n\nThis stuff with the indexing actually gets into the difference between Cardinals (counts) and Ordinals (positions).\n\nSubtraction calculates displacement. Likewise, 0-based indexing implies displacement from the start.\n\nBecause the index is the distance, subtraction works well for Half-Open intervals (include Start, Exclude end, or vice versa).\n\nhttps://aistudio.google.com/prompts/1TgZSYezFfwkXV1mgbMPM5fVqdIzXIt2p","contentHtml":"

This is actually mostly an index hygiene question.

Bounds check first, value check second.

We have to calculate the \"elements removed\" with a - 1 because we don't want to count the destination.

We're looking for the open interval. Therefore, we need to take off the destination.

This... goes pretty deep. I'm not even talking about the algorithm here, which is just a kind of the merge of merge sort, which is just comparing numbers.

This stuff with the indexing actually gets into the difference between Cardinals (counts) and Ordinals (positions).

Subtraction calculates displacement. Likewise, 0-based indexing implies displacement from the start.

Because the index is the distance, subtraction works well for Half-Open intervals (include Start, Exclude end, or vice versa).

https://aistudio.google.com/prompts/1TgZSYezFfwkXV1mgbMPM5fVqdIzXIt2p

\n","code":"$91","codeHtml":"$92"}],"tags":["type:leetcode","topic:array","topic:two-pointers","topic:stack","topic:monotonic-stack","topic:binary-search","difficulty:medium"],"category":"leetcode","createdAt":1765589527664,"updatedAt":1766898121625,"score":"$undefined","review":"$undefined"},{"id":"d92fe8ed-6d08-4bee-9318-4c631ad01540","userId":"$undefined","title":"99. Recover Binary Search Tree","slug":"recover-binary-search-tree","url":"https://leetcode.com/problems/recover-binary-search-tree/","defaultIndex":0,"versions":[{"label":"v1","content":"Find the violation, potentially the second violation (maybe it's globally incorrect and not just locally incorrect), then swap.\n\nBy definition, inorder traversal of a Binary Search Tree gives us a sorted list (recall: Iterative has us go through the entire while loop down the left, and the recursive solution involves us first calling dfs(node.left).\n\nhttps://aistudio.google.com/prompts/153t3kmNldsbAV8b1wl8ETsvBLiWmcIgS","contentHtml":"

Find the violation, potentially the second violation (maybe it's globally incorrect and not just locally incorrect), then swap.

By definition, inorder traversal of a Binary Search Tree gives us a sorted list (recall: Iterative has us go through the entire while loop down the left, and the recursive solution involves us first calling dfs(node.left).

https://aistudio.google.com/prompts/153t3kmNldsbAV8b1wl8ETsvBLiWmcIgS

\n","code":"$93","codeHtml":"$94"},{"label":"v2","content":"Iterative version.","contentHtml":"

Iterative version.

\n","code":"$95","codeHtml":"$96"}],"tags":["type:leetcode","topic:tree","topic:depth-first-search","topic:binary-search-tree","topic:binary-tree","difficulty:medium"],"category":"leetcode","createdAt":1765587025264,"updatedAt":1765587025264,"score":"$undefined","review":"$undefined"},{"id":"db9f1d91-980b-4f8f-824b-a20208e1def5","userId":"$undefined","title":"847. Shortest Path Visiting All Nodes","slug":"shortest-path-visiting-all-nodes","url":"https://leetcode.com/problems/shortest-path-visiting-all-nodes/","defaultIndex":0,"versions":[{"label":"v1","content":"This is relatively pedestrian. Past the bitmasking part, this algorithm is extremely pedestrian. It's just standard BFS.\n\nThough we are not traversing the Node Graph (which has N nodes); we are traversing the State Graph (which has N * 2^N nodes).\n\nSince the constraint is small, it's a giveaway that we're meant to use Bitmask BFS (or Dynamic Programming with Bitmask).\n\nAdditionally, since we can start at any node, we don't simply push (0, mask) into the queue. We initialize the queue with all possible starting nodes.","contentHtml":"

This is relatively pedestrian. Past the bitmasking part, this algorithm is extremely pedestrian. It's just standard BFS.

Though we are not traversing the Node Graph (which has N nodes); we are traversing the State Graph (which has N * 2^N nodes).

Since the constraint is small, it's a giveaway that we're meant to use Bitmask BFS (or Dynamic Programming with Bitmask).

Additionally, since we can start at any node, we don't simply push (0, mask) into the queue. We initialize the queue with all possible starting nodes.

\n","code":"$97","codeHtml":"$98"}],"tags":["type:leetcode","topic:bit-manipulation","topic:breadth-first-search","topic:graph","topic:dynamic-programming","topic:bitmask","difficulty:hard"],"category":"leetcode","createdAt":1765585107195,"updatedAt":1765585107195,"score":"$undefined","review":"$undefined"},{"id":"e4a3715c-fd6a-4e43-af1a-5a57c365b20c","userId":"$undefined","title":"496. Next Greater Element I","slug":"next-greater-element-i","url":"https://leetcode.com/problems/next-greater-element-i/","defaultIndex":0,"versions":[{"label":"v1","content":"$99","contentHtml":"$9a","code":"class Solution:\n def nextGreaterElement(self, nums1, nums2):\n # Dictionary to store the next greater element for each number in nums2\n next_greater_map = {}\n stack = []\n \n # Iterate through nums2 to build the map\n for num in nums2:\n # While stack is not empty and current number is greater than stack top\n while stack and num > stack[-1]:\n val = stack.pop()\n next_greater_map[val] = num\n stack.append(num)\n \n # Iterate through nums1 to build the result\n # If the number exists in map, use the value. Otherwise -1.\n res = []\n for num in nums1:\n if num in next_greater_map:\n res.append(next_greater_map[num])\n else:\n res.append(-1)\n \n return res","codeHtml":"$9b"}],"tags":["type:leetcode","difficulty:easy","topic:array","topic:hash-table","topic:stack","topic:monotonic-stack"],"category":"leetcode","createdAt":1764740212986,"updatedAt":1764740212986,"score":"$undefined","review":"$undefined"},{"id":"8f716cd7-e51f-4065-9542-be54b546dbe8","userId":"$undefined","title":"679. 24 Game","slug":"24-game","url":"https://leetcode.com/problems/24-game/","defaultIndex":0,"versions":[{"label":"v1","content":"$9c","contentHtml":"$9d","code":"$9e","codeHtml":"$9f"}],"tags":["type:leetcode","difficulty:hard","topic:array","topic:math","topic:backtracking"],"category":"leetcode","createdAt":1764738844904,"updatedAt":1764738844904,"score":"$undefined","review":"$undefined"},{"id":"f221d637-ee07-4601-8149-5dbf989f1406","userId":"$undefined","title":"85. Maximal Rectangle","slug":"maximal-rectangle","url":"https://leetcode.com/problems/maximal-rectangle/","defaultIndex":0,"versions":[{"label":"v1","content":"$a0","contentHtml":"$a1","code":"$a2","codeHtml":"$a3"}],"tags":["type:leetcode","difficulty:hard","topic:array","topic:hash-table","topic:dynamic-programming","topic:stack","topic:matrix","topic:monotonic-stack"],"category":"leetcode","createdAt":1764737916103,"updatedAt":1764737916103,"score":"$undefined","review":"$undefined"},{"id":"cc989e73-9d3e-496e-80fc-1970a48fb26d","userId":"$undefined","title":"37. Sudoku Solver","slug":"sudoku-solver","url":"https://leetcode.com/problems/sudoku-solver/","defaultIndex":0,"versions":[{"label":"v1","content":"This is actually quite pedestrian. It's just backtracking.\n\nApproach: Backtracking.\n\nOptimization: Use Sets/HashMaps to make validity checks $O(1)$.\n\nAdvanced (optional mention): Mention that you could optimize by picking the cell with the fewest options (MRV) rather than just the next empty one.","contentHtml":"

This is actually quite pedestrian. It's just backtracking.

Approach: Backtracking.

Optimization: Use Sets/HashMaps to make validity checks $O(1)$.

Advanced (optional mention): Mention that you could optimize by picking the cell with the fewest options (MRV) rather than just the next empty one.

\n","code":"$a4","codeHtml":"$a5"}],"tags":["type:leetcode","difficulty:hard","topic:array","topic:hash-table","topic:backtracking","topic:matrix"],"category":"leetcode","createdAt":1764729667348,"updatedAt":1764729667348,"score":"$undefined","review":"$undefined"},{"id":"26d0342a-351e-4659-b591-c6169ecddf3e","userId":"$undefined","title":"448. Find All Numbers Disappeared in an Array","slug":"find-all-numbers-disappeared-in-an-array","url":"https://leetcode.com/problems/find-all-numbers-disappeared-in-an-array/","defaultIndex":0,"versions":[{"label":"v1","content":"Easy is obviously to just take a set and then iterate through to check what's not in it, but the other way is to treat it like \"Find All Duplicates in an Array,\" which is essentially the inverse of this problem. It uses the exact same trick — using the array indices as a hash map — but you check for the condition at a different time.\n\nhttps://gemini.google.com/app/674ea65b6423a26d","contentHtml":"

Easy is obviously to just take a set and then iterate through to check what's not in it, but the other way is to treat it like \"Find All Duplicates in an Array,\" which is essentially the inverse of this problem. It uses the exact same trick — using the array indices as a hash map — but you check for the condition at a different time.

https://gemini.google.com/app/674ea65b6423a26d

\n","code":"class Solution:\n def findDisappearedNumbers(self, nums):\n # Pass 1: Mark existing numbers by negating values at the corresponding index\n for n in nums:\n # We use abs() because the number might have been negated already by a previous step\n idx = abs(n) - 1\n \n # Only negate if it's currently positive\n if nums[idx] > 0:\n nums[idx] *= -1\n\n # Pass 2: Collect indices that contain positive numbers\n res = []\n for i in range(len(nums)):\n if nums[i] > 0:\n # If the number at index i is positive, i+1 was missing\n res.append(i + 1)\n \n return res","codeHtml":"$a6"}],"tags":["type:leetcode","difficulty:easy","topic:array","topic:hash-table"],"category":"leetcode","createdAt":1764728422079,"updatedAt":1764728422079,"score":"$undefined","review":"$undefined"},{"id":"cc55555e-8beb-47b1-8894-6252f63063ce","userId":"$undefined","title":"240. Search a 2D Matrix II","slug":"search-a-2d-matrix-ii","url":"https://leetcode.com/problems/search-a-2d-matrix-ii/","defaultIndex":0,"versions":[{"label":"v1","content":"We can see it as a binary search tree by starting at the top right (or bottom left)!\n\nhttps://gemini.google.com/app/e1eeb37b7e2489ba","contentHtml":"

We can see it as a binary search tree by starting at the top right (or bottom left)!

https://gemini.google.com/app/e1eeb37b7e2489ba

\n","code":"class Solution:\n def searchMatrix(self, matrix, target):\n if not matrix or not matrix[0]:\n return False\n \n m, n = len(matrix), len(matrix[0])\n \n # Start at Top-Right Corner\n row = 0\n col = n - 1\n \n # Keep going as long as we are inside the matrix boundaries\n while row < m and col >= 0:\n current_val = matrix[row][col]\n \n if current_val == target:\n return True\n \n elif current_val > target:\n # Too big? Eliminate this column, move left\n col -= 1\n \n else: # current_val < target\n # Too small? Eliminate this row, move down\n row += 1\n \n return False","codeHtml":"$a7"}],"tags":["type:leetcode","difficulty:medium","topic:array","topic:binary-search","topic:divide-and-conquer","topic:matrix"],"category":"leetcode","createdAt":1764712801355,"updatedAt":1764728218889,"score":"$undefined","review":"$undefined"},{"id":"8e6dbdbd-1b60-46bf-af9c-aeb292ef9b46","userId":"$undefined","title":"234. Palindrome Linked List","slug":"palindrome-linked-list","url":"https://leetcode.com/problems/palindrome-linked-list/","defaultIndex":0,"versions":[{"label":"v1","content":"You are performing one single destructive act per loop. Everything else is just shuffling your feet to get ready for the next one.\n\n\n\nThe trick is that we don't create new nodes. We just take the existing nodes in the second half of the list and flip their arrows ($next$ pointers) to point backward.\n\nThe Algorithm:\n\nFind the Middle: Use two pointers, Slow (moves 1 step) and Fast (moves 2 steps). When Fast hits the end, Slow is at the middle.\n\nReverse the Second Half: Take the list starting from slow.next and reverse the pointers in-place.\n\nCompare: Run one pointer from the head and one from the tail (the start of the reversed half). If values match, it's a palindrome.\n\n(Optional) Restore: Reverse the second half back to original form so you don't mess up the input list for the caller.","contentHtml":"

You are performing one single destructive act per loop. Everything else is just shuffling your feet to get ready for the next one.

The trick is that we don't create new nodes. We just take the existing nodes in the second half of the list and flip their arrows ($next$ pointers) to point backward.

The Algorithm:

Find the Middle: Use two pointers, Slow (moves 1 step) and Fast (moves 2 steps). When Fast hits the end, Slow is at the middle.

Reverse the Second Half: Take the list starting from slow.next and reverse the pointers in-place.

Compare: Run one pointer from the head and one from the tail (the start of the reversed half). If values match, it's a palindrome.

(Optional) Restore: Reverse the second half back to original form so you don't mess up the input list for the caller.

\n","code":"class Solution:\n def isPalindrome(self, head):\n if not head or not head.next:\n return True\n\n # 1. Find the Middle using Fast/Slow pointers\n slow = head\n fast = head\n while fast and fast.next:\n slow = slow.next\n fast = fast.next.next\n \n # 2. Reverse the second half in-place\n # 'prev' will eventually become the head of the reversed half\n prev = None\n curr = slow # 'slow' is roughly the middle\n while curr:\n next_temp = curr.next\n curr.next = prev\n prev = curr\n curr = next_temp\n \n # 3. Compare the two halves\n left = head\n right = prev # 'prev' is now the start of the reversed right half\n \n while right: # We only check until the right half ends\n if left.val != right.val:\n return False\n left = left.next\n right = right.next\n \n return True","codeHtml":"$a8"},{"label":"v2","content":"","contentHtml":"","code":"class Solution:\n def isPalindrome(self, head):\n vals = []\n current = head\n # 1. Copy to list\n while current:\n vals.append(current.val)\n current = current.next\n \n # 2. Check if list is palindrome\n return vals == vals[::-1]","codeHtml":"$a9"},{"label":"v3","content":"When you look at the code, it looks like 4 lines of chaos. But in reality:\n\nLines 1, 3, and 4 are just admin work. You are updating your local variables (stack variables). You aren't touching the data structure.\n\nLine 2 is the only actual event.\n\nThe \"Surgical\" View\nIf you imagine the Linked List as a long chain of people holding hands:\n\nStep 1: You look at who the current person (curr) is holding hands with on the right (next). You make a mental note.\n\nStep 2 ( The Mutation): You force curr to let go of the right hand and grab the left hand (prev).\n\nStep 3 & 4: You physically step sideways to the next person.\n\nThe \"chain\" isn't exploding. You are just walking down the line, forcing one person at a time to turn around.","contentHtml":"

When you look at the code, it looks like 4 lines of chaos. But in reality:

Lines 1, 3, and 4 are just admin work. You are updating your local variables (stack variables). You aren't touching the data structure.

Line 2 is the only actual event.

The \"Surgical\" View
If you imagine the Linked List as a long chain of people holding hands:

Step 1: You look at who the current person (curr) is holding hands with on the right (next). You make a mental note.

Step 2 ( The Mutation): You force curr to let go of the right hand and grab the left hand (prev).

Step 3 & 4: You physically step sideways to the next person.

The \"chain\" isn't exploding. You are just walking down the line, forcing one person at a time to turn around.

\n","code":"# 1. READ (Safe)\n# We are just looking at the map and remembering where to go next.\n# No structure changes.\nnext_temp = curr.next \n\n# 2. MUTATE (The \"One Thing\")\n# This is the ONLY time we touch the heap/memory structure.\n# We flip exactly one arrow.\ncurr.next = prev\n\n# 3. MOVE (Safe)\n# We just update our own internal label 'prev' to point to the current node.\n# No structure changes.\nprev = curr\n\n# 4. MOVE (Safe)\n# We update our label 'curr' to point to the saved location.\n# No structure changes.\ncurr = next_temp","codeHtml":"$aa"}],"tags":["type:leetcode","difficulty:easy","topic:linked-list","topic:two-pointers","topic:stack","topic:recursion"],"category":"leetcode","createdAt":1764709689197,"updatedAt":1764709709819,"score":"$undefined","review":"$undefined"},{"id":"38ac1758-912d-45fa-9e8d-0bd12e3dc812","userId":"$undefined","title":"103. Binary Tree Zigzag Level Order Traversal","slug":"binary-tree-zigzag-level-order-traversal","url":"https://leetcode.com/problems/binary-tree-zigzag-level-order-traversal/","defaultIndex":0,"versions":[{"label":"v1","content":"The Strategy: Standard Traversal, Alternating Storage\n\nImagine you are walking down the tree level by level.\n\nThe Queue: Always process nodes from Left to Right. This keeps your traversal logic simple and consistent.\n\nThe Result List: This is where you handle the zigzag.\n\nEven Levels (L $\\to$ R): Append values to the end of the level's list.\n\nOdd Levels (R $\\leftarrow$ L): Insert values at the front of the level's list (or just reverse the list at the end).\n\nhttps://gemini.google.com/app/52bb3f3d49d32ff3","contentHtml":"

The Strategy: Standard Traversal, Alternating Storage

Imagine you are walking down the tree level by level.

The Queue: Always process nodes from Left to Right. This keeps your traversal logic simple and consistent.

The Result List: This is where you handle the zigzag.

Even Levels (L $\\to$ R): Append values to the end of the level's list.

Odd Levels (R $\\leftarrow$ L): Insert values at the front of the level's list (or just reverse the list at the end).

https://gemini.google.com/app/52bb3f3d49d32ff3

\n","code":"$ab","codeHtml":"$ac"}],"tags":["type:leetcode","difficulty:medium","topic:tree","topic:binary-tree","topic:breadth-first-search"],"category":"leetcode","createdAt":1764632162328,"updatedAt":1764646622507,"score":"$undefined","review":"$undefined"},{"id":"c7137995-756a-4a8b-a0a6-674b3379c780","userId":"$undefined","title":"214. Shortest Palindrome","slug":"shortest-palindrome","url":"https://leetcode.com/problems/shortest-palindrome/","defaultIndex":0,"versions":[{"label":"v1","content":"https://gemini.google.com/app/73aeca79ed3ba753","contentHtml":"

https://gemini.google.com/app/73aeca79ed3ba753

\n","code":"class Solution:\n def shortestPalindrome(self, s: str) -> str:\n # 1. Reverse the string\n r = s[::-1]\n \n # 2. Iterate through the length of the string\n # We want to find the largest 'i' such that \n # the prefix of s matches the suffix of r\n for i in range(len(s) + 1):\n # s.startswith(r[i:]) checks if the beginning of s \n # matches the reversed string minus the first i characters.\n if s.startswith(r[i:]):\n # The part that didn't match (r[:i]) is exactly \n # the reverse of the suffix we need to add.\n return r[:i] + s\n \n return \"\" # Should basically never hit here given s starts with s","codeHtml":"$ad"},{"label":"v2","content":"","contentHtml":"","code":"$ae","codeHtml":"$af"}],"tags":["type:leetcode","difficulty:hard","topic:string","topic:hash-function","topic:rolling-hash","topic:string-matching"],"category":"leetcode","createdAt":1764631198005,"updatedAt":1764631334262,"score":"$undefined","review":"$undefined"},{"id":"4622d7d7-18ae-4a0a-8ad3-3072e9932d5e","userId":"$undefined","title":"934. Shortest Bridge","slug":"shortest-bridge","url":"https://leetcode.com/problems/shortest-bridge/","defaultIndex":0,"versions":[{"label":"v1","content":"Phase 1: Id