Use the BST property to guide a search toward the target, keeping track of the closest value seen so far. At each node, update the closest if the current node is nearer to the target, then move left if the target is smaller or right if larger. This yields O(h) time and O(1) space.
Pro tip: Explicitly discuss edge cases like an empty tree, duplicate values, and integer overflow when computing differences. Also, mention that an in-order traversal would give a sorted array but is less efficient than the BST-guided approach.
Ask about the tree structure (is it a valid BST?), whether the target can be outside the range of values, and if there are duplicate values. Confirm the expected return type (node value, not node).
Explain that you'll traverse from the root, maintaining the closest value found so far. At each node, compare the absolute difference between node value and target, updating the closest if needed.
Describe how to decide the direction: if target < node value, go left; if target > node value, go right; if equal, return immediately. This leverages the BST property to prune the search space.
State that time complexity is O(h) where h is the tree height (O(log n) for balanced, O(n) worst-case), and space is O(1) for iterative. Mention handling empty tree, single node, and target outside range.
Briefly mention that an in-order traversal to get a sorted list and then binary search would be O(n) time and O(n) space, which is less efficient. Also, note that if the tree is not a BST, a full traversal is needed.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
The base problem I handled fine, classic stack approach to track unmatched indices then rebuild the string.
Start by clarifying the problem and edge cases, then propose a stack-based solution that tracks indices of unmatched parentheses. For the follow-up, explain a two-pass counting method that uses O(1) space by first removing invalid closing parentheses and then invalid opening ones.
Pro tip: Meta interviewers value clean, efficient code and clear communication. Before coding, discuss trade-offs between the stack and two-pass approaches, and mention that the two-pass method is optimal for space but requires careful handling of indices.
Ask if the input can contain other characters, if the output should be any valid string or a specific one, and confirm that we only remove parentheses.
Use a stack to store indices of unmatched opening parentheses. When a closing parenthesis is encountered, pop if possible; otherwise, mark it for removal. After traversal, mark remaining stack indices for removal.
First pass: count and remove invalid closing parentheses by tracking balance. Second pass (right-to-left): remove invalid opening parentheses similarly. This uses only counters and modifies the string in place.
Compare time and space: stack uses O(n) space, two-pass uses O(1) extra space but may require multiple passes. Discuss when each is preferable.
Walk through examples like '(()', ')()', and 'a)b(c' to verify correctness and edge cases.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.