BFS level-order traversal, grab the last node at each level.
Use BFS level-order traversal, recording the last node's value at each level. Alternatively, use DFS prioritizing the right child, tracking depth to capture the first node seen at each new depth. Both approaches yield O(n) time and O(h) space (where h is tree height).
Pro tip: Clarify with the interviewer whether the tree can be empty or skewed, and mention that BFS is more intuitive for level-based problems while DFS uses less memory for balanced trees. Also, discuss how you would handle very large trees where recursion depth might be an issue.
Confirm that 'visible from the right side' means the rightmost node at each depth. Clarify edge cases: empty tree, single node, skewed tree.
Decide between BFS (level-order) and DFS (right-first). Explain the trade-offs: BFS uses a queue and processes level by level; DFS uses recursion/stack and tracks depth.
For BFS: enqueue root, for each level record the last node's value. For DFS: traverse right child first, and when visiting a new depth, add the node's value to the result.
Walk through a sample tree (e.g., [1,2,3,null,5,null,4]) and verify the output. Also test edge cases like empty tree and left-skewed tree.
State time complexity O(n) and space complexity O(h) for DFS or O(w) for BFS (w = max width). Discuss which is more efficient in different scenarios.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
This one tripped me up more than I expected.
Use a stack-based approach to track unmatched parentheses: first pass marks invalid closing parentheses, second pass marks invalid opening parentheses. Then build the result by including only characters that are not marked for removal. Alternatively, use a counter-based method to achieve the same in two passes.
Pro tip: Clarify that 'minimum removal' means removing only unmatched parentheses, and that multiple valid answers may exist—returning any is acceptable. Also mention that the solution runs in O(n) time and O(n) space, which is optimal.
Confirm that the input contains lowercase letters and parentheses, and that you need to remove the minimum number of parentheses to make the string valid. A valid string has balanced parentheses and no unmatched ones.
Decide between a stack-based method (to track indices of unmatched parentheses) or a two-pass counter method (to count unmatched closing and opening parentheses). Both are O(n) time and O(n) space.
In the first pass, mark or count unmatched closing parentheses. In the second pass (if using stack, after reversing or using another stack), mark unmatched opening parentheses.
Build the output string by including only characters that are not marked for removal. Ensure the result is valid and has minimum removals.
Walk through examples like 'a)b(c)d' -> 'ab(c)d' and 'lee(t(c)o)de)' -> 'lee(t(c)o)de' 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.