I got the two-pass stack approach pretty fast, track unmatched opens and then sweep for unmatched closes.
Use a stack-based two-pass approach: first, scan left-to-right to mark unmatched closing parentheses; then scan right-to-left to mark unmatched opening parentheses. Remove all marked parentheses to produce a valid string in O(n) time. Argue minimality by showing each removal is necessary and sufficient.
Pro tip: Emphasize that the algorithm removes exactly the parentheses that cannot be matched, and mention that the two-pass method avoids the need for a stack, achieving O(1) extra space if we use a counter instead of a stack.
Confirm that a valid string has balanced parentheses and that we can return any valid result. Define the problem as removing the minimum number of parentheses to achieve balance.
Use a counter to track balance in a left-to-right pass, marking unmatched closing parentheses. Then do a right-to-left pass to mark unmatched opening parentheses. Remove all marked parentheses.
Explain that each marked parenthesis is part of an unmatched pair or is unmatched itself, so it must be removed. Removing all marked parentheses yields a valid string, and no fewer removals can fix the imbalances.
State that the algorithm runs in O(n) time with two passes, and uses O(n) space if marking with a boolean array, or O(1) extra space if using counters and building the result on the fly.
Walk through an example like 'a)b(c)d' to demonstrate the process. Mention edge cases: empty string, all parentheses, no parentheses, and strings with only opening or only closing parentheses.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
First, clarify that the original problem likely asked for one valid result, and now we need to enumerate all. Then, describe how to modify the algorithm to collect all solutions, such as using backtracking with pruning or BFS with state tracking, ensuring we only explore paths that can still achieve the minimum removals. Finally, discuss trade-offs like increased time and space complexity and how to handle duplicates.
Pro tip: Mention that you would deduplicate results using a set or by sorting and skipping duplicates, and emphasize that the core logic remains the same but the search space expands. This shows you think about efficiency and correctness.
Confirm that the goal is to output all valid strings with the minimum number of removals, and discuss constraints like input size, character set, and whether duplicates should be removed.
Recall the original approach (e.g., BFS level-by-level, DP, or two-pointer) and determine how to adapt it to collect all solutions instead of stopping at the first.
For BFS, process all nodes at the current level and collect valid results; for backtracking, explore all branches that can still reach the minimum removals, using pruning to avoid unnecessary paths.
Use a set to store results to avoid duplicates, and consider memoization or visited states to prevent redundant work, especially if the same substring can be reached via different removal sequences.
Discuss how the time and space complexity increase (potentially exponential in the worst case) and suggest optimizations like early termination or limiting output size if needed.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.