My first instinct was to just scan left to right and track unmatched parens with a counter, but then I realized I also needed to reconstruct the actual string, not just count removals.
Use a stack to track unmatched opening parentheses and a counter for unmatched closing parentheses. After one pass, remove the unmatched closing parentheses and the unmatched opening parentheses (by marking their indices) to produce the valid string. Return the total count of removals and the resulting string.
Pro tip: Clarify whether you need to remove the minimum number of parentheses or if any valid string is acceptable; the stack approach guarantees minimal removals. Also, consider edge cases like empty string or string with no parentheses.
Confirm that you need to remove the fewest parentheses to make the string valid, and return both the count and the resulting string. Ask clarifying questions about input constraints and expected output format.
Use a stack to keep track of indices of unmatched opening parentheses. Also maintain a set or boolean array to mark characters to be removed.
Iterate through the string. For each '(', push its index onto the stack. For each ')', if the stack is not empty, pop an index (matching pair); otherwise, mark this ')' for removal. After the loop, mark all indices remaining in the stack for removal.
Construct the resulting string by including only characters whose indices are not marked for removal. The number of removals is the size of the marked set.
State that the time complexity is O(n) and space complexity is O(n). Test with examples like '(()', ')()', '())(', and strings with no parentheses.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.