I recognized it as a variant of the minimum-remove parentheses problem but spent a few seconds second-guessing myself because they only wanted a count, not the reconstructed string.
Use a single-pass stack-based algorithm: iterate through the string, push open parentheses onto a stack, and for each close parenthesis, pop if the stack is non-empty; otherwise, increment a removal counter. After the pass, add the stack size to the removal counter to account for unmatched open parentheses. This yields the minimum removals in O(n) time and O(n) space.
Pro tip: Clarify that the goal is to return the minimum number of removals, not the resulting valid string. Mention that the stack approach is optimal and can be optimized to O(1) space by using a counter instead of a stack, since only the count of unmatched opens matters.
Restate the problem: given a string with lowercase letters and parentheses, find the minimum number of parentheses to remove so that the remaining parentheses are balanced and properly nested. Confirm that only parentheses matter; letters can be ignored.
Use a stack to track unmatched open parentheses. Alternatively, use a counter for unmatched opens to achieve O(1) space, since we only need the count, not the positions.
Iterate through each character: if it's '(', push to stack or increment open counter; if it's ')', pop or decrement if open counter > 0, else increment removal counter. Ignore other characters.
After traversal, any remaining open parentheses in the stack (or the open counter) are unmatched and must be removed. Add that count to the removal counter.
The sum of unmatched close parentheses (counted during traversal) and unmatched open parentheses (leftover) is the minimum number of removals. Return this integer.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.