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Meta·Software Engineer·Technical Phone Screen·Intermediate

Intermediate
May 2026

Summary

Meta SWE coding round, one question on string manipulation with parentheses. Pretty standard algorithmic problem but the details matter more than you'd think.

Questions Asked (1)

Q1

Given a string of characters containing only '(' and ')', remove the fewest parentheses possible to make the string valid. Return any valid result.

Algorithms & Data Structures
Author's notes

My first instinct was to just scan and count mismatches, but getting the actual removal right took a bit more work.

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AI HintsAI Generated

Suggested Approach

Use a stack or counter to track unmatched parentheses, then remove them. Alternatively, compute the number of invalid open and close parentheses and filter them out. Return any valid string with minimal removals.

Pro tip: Clarify that multiple valid answers exist and that your solution returns one of them; also mention that the algorithm runs in O(n) time and O(n) space, which is optimal.

1. Understand the problem

Confirm that the goal is to remove the fewest parentheses to make the string valid, and that any valid result is acceptable. Clarify that a valid string has balanced parentheses and no unmatched ones.

2. Choose an approach

Decide between using a stack to track indices of unmatched parentheses or using two passes with counters. Both are O(n) time; the stack uses O(n) space, while counters can be O(1) space if only counts are needed.

3. Identify invalid parentheses

Scan left to right to find unmatched closing parentheses (when count of ')' exceeds '('). Then scan right to left to find unmatched opening parentheses. Mark these indices for removal.

4. Construct the result

Build the output string by including only characters whose indices are not marked for removal. Ensure the result is valid and has minimal removals.

5. Analyze complexity and edge cases

State that the algorithm runs in O(n) time and O(n) space (or O(1) if using counters and building result on the fly). Discuss edge cases like empty string, all invalid, or already valid.

Key Points to Mention

  • The problem allows any valid result, so multiple solutions exist.
  • Use a stack to track indices of unmatched parentheses for easy removal.
  • Alternatively, use two passes with counters to identify invalid parentheses.
  • Time complexity is O(n) and space complexity can be O(n) or O(1) depending on implementation.
  • Edge cases: empty string, string with all '(' or all ')', already valid string.
  • Minimal removals are guaranteed by removing only unmatched parentheses.

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