Meta·Machine Learning Engineer·Technical Phone Screen·Senior

Senior

Apr 2026

Summary

Meta MLE interview with a coding question on string manipulation. Pretty standard algorithmic problem but the edge cases are where it gets tricky.

Questions Asked (1)

Given a string containing parentheses and lowercase letters, remove the minimum number of parentheses to make the string valid. Return any valid result.

Algorithms & Data Structures

Author's notes

I went straight to a stack-based approach and it worked, but I fumbled explaining the minimum removal part.

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AI HintsAI Generated

Suggested Approach

Use a stack to track unmatched opening parentheses and a set to mark indices of invalid closing parentheses. Then build the result by skipping marked indices and any unmatched opening parentheses. This ensures the minimum number of removals.

Pro tip: After solving, discuss how you would test edge cases like empty string, all invalid, and nested valid parentheses. Also, mention that the solution runs in O(n) time and space, which is optimal.

1. Clarify the problem

Confirm that the goal is to remove the minimum number of parentheses to make the string valid, and that any valid result is acceptable. Ask if the string can be empty or contain only parentheses.

2. Choose the right data structures

Use a stack to keep track of indices of unmatched opening parentheses. Use a boolean array or set to mark indices of invalid closing parentheses.

3. Traverse and mark invalid parentheses

Iterate through the string: push indices of '(' onto the stack; for ')', if stack is not empty, pop, else mark this index as invalid. After traversal, mark all indices remaining in the stack as invalid.

4. Build the result string

Construct the result by including only characters whose indices are not marked as invalid. Return the resulting string.

5. Analyze complexity and test

State that the time complexity is O(n) and space complexity is O(n). Walk through a few test cases to verify correctness.

Key Points to Mention

Use a stack to track unmatched opening parentheses.
Mark invalid closing parentheses when the stack is empty.
After traversal, mark any remaining opening parentheses as invalid.
Build the result by skipping marked indices.
Time and space complexity are both O(n).
The solution guarantees the minimum number of removals.

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