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Meta·Software Engineer·Technical Phone Screen·Intermediate

Intermediate
Jun 2026

Summary

Meta SWE interview with a valid parentheses problem. Pretty standard coding round, nothing that would surprise anyone who's done leetcode prep.

Questions Asked (1)

Q1

Given a string of parentheses (and possibly other bracket types), determine whether the string is valid, meaning every opening bracket has a corresponding closing bracket in the correct order.

Algorithms & Data Structures
Author's notes

Stack-based solution, pretty much what you'd expect.

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AI HintsAI Generated

Suggested Approach

Start by clarifying the problem: which bracket types are included, and whether the string may contain other characters. Then propose a stack-based solution that pushes opening brackets and pops when a matching closing bracket is encountered, returning false if a mismatch or leftover opening brackets exist.

Pro tip: Mention edge cases like empty string, single bracket, and strings with non-bracket characters, and discuss the time and space complexity (O(n) time, O(n) space) to show thoroughness. Also, briefly note that a counter-based approach works only for a single bracket type, but a stack is needed for multiple types.

1. Clarify requirements

Ask the interviewer which bracket types to consider (e.g., (), [], {}) and whether the string can contain other characters. Confirm that an empty string is considered valid.

2. Choose data structure

Explain that a stack is ideal because brackets must be closed in LIFO order. Mention that a simple counter works only for one bracket type, but a stack handles multiple types.

3. Outline algorithm

Iterate through each character: if it's an opening bracket, push it; if it's a closing bracket, check if the stack is empty or the top doesn't match, then return false; otherwise pop. After the loop, return true only if the stack is empty.

4. Analyze complexity

State that the time complexity is O(n) because each character is processed once, and space complexity is O(n) in the worst case (e.g., all opening brackets).

5. Test with examples

Walk through examples like '()[]{}', '([)]', and '{[]}' to demonstrate correctness and edge cases. Mention that non-bracket characters can be ignored or handled based on clarification.

Key Points to Mention

  • Use a stack to track opening brackets and ensure proper nesting.
  • Handle mismatched closing brackets by checking the top of the stack.
  • After processing, the stack must be empty for the string to be valid.
  • Time complexity O(n) and space complexity O(n).
  • Edge cases: empty string, single bracket, unbalanced brackets, and strings with non-bracket characters.
  • Alternative approach: counter for single bracket type, but stack is necessary for multiple types.

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