My first instinct was to just count opens and closes, which obviously doesn't handle nesting correctly.
Use a stack to track unmatched opening parentheses and a counter for unmatched closing parentheses. After one pass, the minimum removals equals the number of unmatched closing parentheses plus the number of unmatched opening parentheses left in the stack.
Pro tip: Clarify whether you need to return the count or the resulting balanced string; if the latter, you can build the result by marking indices to remove. Also, mention that the greedy stack approach is optimal because each removal fixes exactly one imbalance.
Ask whether the function should return the minimum number of removals or the balanced string itself. Discuss edge cases: empty string, no parentheses, already balanced, all opening or all closing parentheses.
Use a stack to track indices of unmatched opening parentheses. Alternatively, use a counter for unmatched closing parentheses and a stack for unmatched opening parentheses.
Iterate through the string: push opening parentheses onto the stack; for closing parentheses, pop if stack is non-empty, otherwise increment a counter for unmatched closing parentheses.
After the pass, the minimum removals is the sum of the counter for unmatched closing parentheses and the size of the stack (unmatched opening parentheses).
State that time and space complexity are O(n). If returning the string, explain how to mark indices for removal and build the result in O(n) time.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.