My first instinct was to just count opens and closes and subtract, which is wrong.
Use a single-pass counter approach: iterate through the string, increment a counter for each '(', decrement for each ')' but only if the counter is positive; otherwise, count an unmatched ')'. After the pass, the counter holds unmatched '(' and the separate count holds unmatched ')'; sum them for the answer.
Pro tip: Clarify that the problem asks for the minimum number of removals, which equals the total unmatched parentheses. Mention that a stack-based solution also works but uses O(n) space, while the counter approach uses O(1) space—demonstrating awareness of trade-offs.
Confirm that we need to count unmatched parentheses, not remove them. Ensure that other characters are ignored.
Decide between stack-based (O(n) space) and counter-based (O(1) space) solutions. Explain why counter is better for space efficiency.
Initialize open=0, close=0. For each char: if '(', open++; if ')', if open>0 then open-- else close++. Return open+close.
State that time complexity is O(n) and space complexity is O(1).
Validate with edge cases: empty string, all opens, all closes, mixed like '())(', and strings with other characters.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.