Meta·Software Engineer·Technical Phone Screen·Intermediate

IntermediatePrefer not to say

Apr 2026

Summary

Meta coding round, one question the whole time. It's a parentheses problem dressed up slightly differently from the classic LC version, and the stack approach clicks pretty fast once you stop overthinking it.

Questions Asked (1)

Given a string of '(' and ')' characters, remove all matched valid pairs and return the count of unmatched parentheses that remain. Solve it in O(n) time.

Algorithms & Data Structures

Author's notes

I recognized it as a cousin of the longest valid parentheses problem but the ask is subtly different, you're counting what's left over rather than what's valid.

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AI HintsAI Generated

Suggested Approach

Use a single pass with a counter to track unmatched opening parentheses, incrementing on '(' and decrementing on ')' when possible, while counting unmatched closing parentheses separately. At the end, sum the unmatched opening and closing counts to get the total unmatched parentheses.

Pro tip: Clarify with the interviewer whether the count should include both unmatched opening and closing parentheses, and mention that the algorithm can be easily adapted to return the actual unmatched characters if needed.

1. Clarify the problem

Confirm that the goal is to count the total number of unmatched parentheses after removing all valid pairs, and that the input string consists only of '(' and ')'.

2. Choose the right data structure

Decide between using a stack or a simple counter. For this problem, a counter is sufficient and more space-efficient, achieving O(1) space.

3. Design the algorithm

Iterate through the string: for each '(', increment a counter for unmatched openings; for each ')', if there is an unmatched opening, decrement it, otherwise increment a counter for unmatched closings.

4. Implement and test

Write clean code with meaningful variable names, and test with edge cases like empty string, all opening, all closing, and mixed patterns.

5. Analyze complexity

State that the time complexity is O(n) and space complexity is O(1), which is optimal for this problem.

Key Points to Mention

Time complexity O(n) and space complexity O(1) with a counter approach.
Handling of edge cases: empty string, string with only '(' or only ')', and already balanced strings.
The algorithm correctly counts unmatched parentheses by tracking unmatched openings and closings separately.
Alternative stack-based solution and its O(n) space complexity, but note that counter is more efficient.
The importance of clarifying whether to return the count or the actual unmatched characters.
Potential follow-up: modify to return the indices or the string after removing matched pairs.

AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.