My first instinct was to just count unmatched parens with a stack and then do a second pass to remove them.
Use a stack to track indices of unmatched parentheses, then build the result by skipping those indices. Alternatively, use a counter-based two-pass approach to mark invalid parentheses. Both methods run in O(n) time and O(n) space.
Pro tip: Clarify that multiple valid answers exist and any is acceptable, then discuss trade-offs between stack (simpler) and counter (O(1) space) approaches. Mention that the counter method requires two passes but is more space-efficient.
Confirm that we need to remove the minimum number of parentheses to make the string valid, and that any valid result is acceptable. Note that the string contains only '(', ')', and lowercase letters.
Decide between stack-based (track indices of unmatched parentheses) or counter-based (two-pass with balance counters) methods. Explain the trade-offs in time and space complexity.
For stack: iterate through string, push '(' indices, pop on matching ')', and mark unmatched indices. For counter: first pass left-to-right to remove excess ')', second pass right-to-left to remove excess '('.
Build the output string by including only characters whose indices are not marked for removal. Return the resulting valid string.
Walk through examples like 'a)b(c)d' -> 'ab(c)d' and '))((' -> '' to verify correctness. Discuss edge cases such as empty string or all parentheses.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.