Clarify the problem and constraints, then propose a stack-based solution that tracks unmatched parentheses. Explain that you can remove invalid parentheses in a single pass by marking indices to delete, and discuss time/space complexity.
Pro tip: Mention that the problem can be solved in O(n) time with O(n) space using a stack, but also note that a two-pass counting approach can achieve O(1) space if only the count of removals is needed. This shows awareness of optimization trade-offs.
Ask clarifying questions: Are there multiple valid answers? Should we return the string or just the minimum number? Are there other characters besides parentheses and letters? Confirm that letters are ignored and only parentheses matter.
Propose using a stack to track indices of unmatched opening parentheses. Iterate through the string, push indices of '(' and pop for ')' when possible; after traversal, mark remaining stack indices and any unmatched ')' for removal.
Trace the algorithm on a sample input like 'a)b(c)d' to demonstrate how invalid parentheses are identified and removed, resulting in 'ab(c)d'.
State that the algorithm runs in O(n) time and uses O(n) space for the stack and a boolean array to mark removals. Mention that space can be reduced to O(1) if only the count is needed.
Cover edge cases: empty string, string with only letters, all opening or all closing parentheses, and nested valid parentheses. Explain how the algorithm handles each.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Trickier than it looks because you can go up the tree too, not just down.
Treat the tree as an undirected graph and perform a BFS from the target node, tracking distance until K edges are reached. Alternatively, use a two-phase DFS: first find the path from root to target, then for each ancestor at distance d from target, explore the subtree of the other child to a depth of K-d-1. Both approaches run in O(N) time.
Pro tip: Clarify whether the tree is a binary tree (each node has at most two children) and whether nodes are unique. Also, mention that if K=0, the answer is just the target node itself, and if K exceeds the tree height, the result is empty.
Confirm the definition of 'K edges away' and whether the tree is rooted or unrooted. Ask if nodes are unique and if the target node is guaranteed to exist.
Decide between BFS on the undirected graph (simpler, uses extra space for parent pointers) or DFS with path tracking (more space-efficient, but requires careful implementation).
For BFS: build parent pointers, then BFS from target, stopping at distance K. For DFS: find the path from root to target, then for each ancestor, explore the other subtree to depth K - distance - 1.
State that both approaches run in O(N) time and O(N) space in the worst case (for BFS, space is O(N) for the queue and parent map; for DFS, space is O(H) for recursion stack plus O(H) for path).
Consider K=0, K larger than tree height, target at root, target at leaf, and skewed trees. Verify that the solution handles these correctly.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.