I went with the linear scan approach, two counters tracking unmatched closing parens and currently open ones, sum them at the end.
Use a single-pass counter approach: track the number of unmatched closing parentheses and the current open balance. For each character, update these counters; at the end, the minimum insertions equals unmatched closings plus remaining open balance.
Pro tip: Clarify that the solution runs in O(n) time and O(1) space, and mention that this is optimal since you must examine each character at least once. Also, briefly discuss how the approach handles edge cases like empty strings or already valid strings.
Confirm that the string contains only '(' and ')' and that insertions can be made anywhere. Ask if the goal is to return the minimum number, not the resulting string.
Initialize two counters: 'open' for unmatched '(' and 'insertions' for unmatched ')'. Iterate through the string.
If char is '(', increment open. If char is ')', check if open > 0; if so, decrement open (match), else increment insertions (need a '(' before).
After the loop, the minimum insertions needed is insertions + open. Return that sum.
State that time complexity is O(n) and space complexity is O(1). Mention that this is optimal.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.