Use Floyd's cycle-finding algorithm (tortoise and hare) to detect a cycle and find the meeting point. Then, reset one pointer to the head and move both pointers one step at a time until they meet again; that node is the start of the cycle. If no cycle is detected, return null.
Pro tip: Clearly explain why the algorithm works, especially the mathematical proof that the distance from the head to the cycle start equals the distance from the meeting point to the cycle start (modulo the cycle length). This demonstrates deep understanding and is often expected at Meta.
Initialize two pointers, slow and fast, at the head. Move slow one step and fast two steps at a time until they meet or fast reaches null. If fast reaches null, there is no cycle; return null.
If a cycle exists, the slow and fast pointers will meet at some node inside the cycle. Record this meeting node.
Reset one pointer to the head while keeping the other at the meeting point. Move both pointers one step at a time until they meet again. The node where they meet is the start of the cycle.
Return the node where the cycle begins, or null if no cycle was detected.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Two cases: node has a right subtree, or it doesn't.
Clarify that the BST is not necessarily balanced and that parent pointers are available. Then describe the two-case algorithm: if the node has a right subtree, return the leftmost node in that subtree; otherwise, walk up parent pointers until you find an ancestor for which the node lies in its left subtree. Analyze time and space complexity, noting O(h) time and O(1) space.
Pro tip: Explicitly state that the algorithm uses O(1) extra space and runs in O(h) time, where h is the tree height, and mention that this is optimal for the given structure. Also, briefly note that if the tree were balanced, h = O(log n), but in the worst case (skewed tree) it's O(n).
Confirm that the BST may be unbalanced and that parent pointers are valid. Discuss edge cases: node is null, node has no successor (e.g., maximum node), and tree with a single node.
If the given node has a right child, the in-order successor is the leftmost node in its right subtree. Explain how to find it by traversing left pointers until null.
If there is no right child, walk up using parent pointers. While the current node is a right child of its parent, keep moving up. The successor is the first ancestor for which the node is in the left subtree, or null if none exists.
State that time complexity is O(h) and space is O(1). Walk through a small example to verify correctness, including cases where the successor is an ancestor or does not exist.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
I started with the naive 'scan the row and column after each move' approach and they immediately asked about the time complexity.
Start by clarifying the requirements: n×n board, move operation that places a mark and returns whether that player wins, O(1) time per move. Then design a class that maintains row, column, and diagonal counts for each player, updating them on each move and checking for a win in constant time. Finally, discuss edge cases and potential optimizations.
Pro tip: Mention that the O(1) win check is achieved by tracking counts per row, column, and diagonal for each player, and that this approach scales to any n. Also, note that you can avoid storing the entire board if you only need to detect wins, but storing it may be useful for other operations like undo or display.
Confirm the board size n, the move operation signature, and that each move must be O(1). Ask about win conditions (e.g., only the player who just moved can win) and whether the board needs to be stored.
Use arrays to track counts for each player: rowCounts[player][row], colCounts[player][col], and two variables for diagonals. Optionally, store the board as a 2D array for completeness.
On move(row, col, player), update the corresponding row, column, and diagonal counts. Check if any count reaches n; if so, return true, else false. Ensure O(1) time by only updating relevant counters.
Consider invalid moves (out of bounds, occupied cell), multiple wins, and whether to support undo. Discuss space-time trade-offs: O(n) space for counts vs O(n^2) for board.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Label each island with a unique ID and compute its size using BFS/DFS, then for each water cell, sum the sizes of adjacent distinct islands plus one. The maximum of these sums is the answer, and if no water exists, return the largest island size.
Pro tip: Emphasize that the solution must be O(n^2) time and space to handle 500x500 grids, and mention that using a hash set to deduplicate adjacent island IDs avoids double-counting.
Confirm grid size up to 500x500, binary values, and that flipping is optional. Discuss edge cases: no water, all water, and multiple islands.
Traverse the grid with BFS/DFS to assign each land cell an island ID and record the size of each island in a map or array.
For each 0, collect the unique IDs of adjacent islands, sum their sizes, and add 1 for the flipped cell. Track the maximum.
If no water cells exist, return the size of the largest island (or 0 if grid is all water).
Explain that the algorithm runs in O(n^2) time and uses O(n^2) space for the ID grid and size map, which is optimal for this problem.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.