Stack-based approach, track indices of unmatched parens and remove them at the end.
Clarify the problem: we need to remove the minimum number of parentheses to make the string valid, meaning every opening parenthesis has a matching closing parenthesis in the correct order. Use a stack to track unmatched opening parentheses and a set to mark unmatched closing parentheses, then build the result by skipping marked indices. Alternatively, use two passes to count and remove invalid parentheses.
Pro tip: After presenting the stack solution, mention that a two-pass counting approach can achieve O(1) space if the output can be built in place or if we only need the length. This shows awareness of space optimization, which is valued at Meta.
Confirm that we need to remove the minimum number of parentheses to make the string valid, and that the relative order of remaining characters is preserved. Discuss edge cases: empty string, already valid string, string with only invalid parentheses, and multiple valid solutions.
Decide between a stack-based solution (O(n) time, O(n) space) and a two-pass counting solution (O(n) time, O(1) space). Explain the trade-offs and pick one to implement, typically the stack approach for clarity.
For the stack approach: iterate through the string, push indices of '(' onto a stack, and for ')' either pop if stack is non-empty or mark the index as invalid. After the first pass, mark all indices left in the stack as invalid. Then build the result by skipping marked indices.
Walk through examples like '(()', ')()', '()())', and 'a)b(c' to verify correctness. Ensure the output is valid and has minimum removals.
State time and space complexity: O(n) time and O(n) space for the stack approach. Mention that a two-pass counting method can reduce space to O(1) if we only need the length or can modify the string in place.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
This one took me a minute to get the right approach.
First, identify all existing islands and label each cell with a unique island ID, recording the size of each island. Then, for each water cell (0), compute the sum of sizes of distinct neighboring islands plus one, and track the maximum. If no water cells exist, return the size of the largest island.
Pro tip: Clarify edge cases upfront, such as when the grid is all 1s or all 0s, and discuss the trade-offs between DFS/BFS and Union-Find, showing you consider both correctness and efficiency.
Confirm the problem constraints: grid dimensions, whether flipping is mandatory, and what to return if no 0 exists. Restate the goal to ensure alignment.
Use DFS, BFS, or Union-Find to traverse the grid, assign a unique ID to each island, and record its size in a map or array.
For every 0, look at its four neighbors, collect the distinct island IDs, sum their sizes, add 1 for the flipped cell, and update the maximum.
If there are no 0s, return the size of the largest existing island. If the grid is all 0s, flipping one cell yields an island of size 1.
State that the time and space complexity are O(R*C) for an R x C grid, as each cell is visited a constant number of times.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Clarify that the problem asks for the rightmost node at each level, then choose between BFS (level-order traversal) and DFS (preorder traversal with depth tracking). For BFS, process each level and record the last node; for DFS, track the maximum depth seen so far and update the result when visiting a node at a new depth, prioritizing the right child first.
Pro tip: At Meta, interviewers value clean, efficient code and clear communication. Start by discussing the trade-offs between BFS and DFS (e.g., BFS uses O(width) space, DFS uses O(height) space) and pick the one that best fits the constraints, then write modular code with meaningful variable names.
Confirm that the output should be a list of values from the rightmost node at each level, ordered from top to bottom. Ask about edge cases like an empty tree or a tree with only left children.
Decide between BFS and DFS based on space complexity and simplicity. BFS is straightforward: use a queue and process level by level. DFS is more memory-efficient for skewed trees: use recursion with depth tracking.
For BFS: initialize a queue with the root, while the queue is not empty, iterate over the current level size, and add the last node's value to the result. For DFS: recursively traverse right child first, and if the current depth equals the result size, append the node's value.
Walk through a sample tree (e.g., [1,2,3,null,5,null,4]) and verify the output matches expectations. Also test edge cases: empty tree, single node, and a left-skewed tree.
State the time complexity: O(n) for both approaches since each node is visited once. Space complexity: O(w) for BFS where w is the maximum width, and O(h) for DFS where h is the height (due to recursion stack).
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.