The trick is you don't have the root, so you can't do the usual recursive thing.
Since there is no root reference, use the parent pointers to traverse upward from each node to the root, storing visited nodes in a hash set. Then traverse from the other node upward until a node is found in the set; that node is the LCA. Alternatively, compute the depths by traversing to the root, align the depths, and move both pointers upward until they meet.
Pro tip: Clarify with the interviewer whether modifying the tree is allowed and whether the nodes are guaranteed to be in the same tree. Also, mention that the hash set approach uses O(h) space, but if space is a concern, the depth alignment method uses O(1) extra space.
Ask if the two nodes are guaranteed to be in the same tree, if parent pointers are valid, and if we can modify the tree. Also, discuss space and time constraints.
Decide between the hash set method (simpler, O(h) space) and the depth alignment method (O(1) space). Explain the trade-offs.
For hash set: traverse from node1 to root, storing each node in a set; then traverse from node2 upward until a node is in the set. For depth alignment: compute depths by traversing to root, align depths, then move both pointers until they meet.
State time complexity O(h) where h is the height of the tree, and space complexity O(h) for hash set or O(1) for depth alignment.
Consider cases where one node is an ancestor of the other, nodes are the same, or the tree is skewed. Verify the algorithm handles them correctly.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Clarify that the goal is to remove the minimum number of parentheses to make the string valid, meaning every closing parenthesis must have a matching opening parenthesis before it. Use a stack to track unmatched opening parentheses and a set to mark unmatched closing parentheses, then build the result by skipping marked characters. Alternatively, use a two-pass counting method to identify invalid parentheses.
Pro tip: Mention that the two-pass counting method uses O(1) space and is more efficient than a stack, but the stack approach is easier to explain and code under pressure. Also, discuss how to handle multiple valid outputs by noting that any valid string with minimum removals is acceptable.
Restate the problem: remove the minimum number of parentheses to make the string valid, where valid means every closing parenthesis has a matching opening parenthesis before it. Confirm that letters are ignored and only parentheses matter.
Decide between stack-based and two-pass counting methods. Explain the trade-offs: stack uses O(n) space but is intuitive; two-pass uses O(1) space but requires careful counting.
For stack: iterate through string, push indices of '(' onto stack, pop on matching ')', and mark unmatched ')' indices. After iteration, mark remaining '(' indices. For two-pass: first pass left-to-right counts open and marks invalid ')'; second pass right-to-left counts close and marks invalid '('.
Construct the output string by including only characters whose indices are not marked for removal. Ensure the order of characters is preserved.
State time complexity O(n) and space complexity O(n) for stack or O(1) for two-pass. Walk through edge cases: empty string, all parentheses, no parentheses, nested and sequential parentheses.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.