Bloomberg·Software Engineer·Technical Phone Screen·Intermediate

Intermediate

Apr 2026

Summary

Bloomberg SWE coding round, one question on string manipulation with parentheses. Pretty standard algorithmic problem but the stack angle is easy to miss if you haven't seen it before.

Questions Asked (1)

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

Algorithms & Data Structures

Author's notes

My first instinct was to just scan left to right and count unmatched parens, which kind of works but gets messy fast.

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AI HintsAI Generated

Suggested Approach

Use a stack-based approach to identify unmatched parentheses in a single pass, then remove them. Alternatively, use a counter-based method to track balance and mark invalid parentheses for removal. Return the string after removing marked characters.

Pro tip: Discuss the trade-offs between the stack and counter approaches, and mention that the counter method uses O(1) space. Also, clarify that any valid result is acceptable, so you don't need to find all possible solutions.

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 if there are constraints on time/space.

2. Choose an approach

Decide between a stack-based method (which uses O(n) space) and a counter-based method (which uses O(1) space). Explain the trade-offs and pick one to implement.

3. Identify invalid parentheses

Traverse the string to find unmatched closing parentheses (when balance is negative) and unmatched opening parentheses (leftover balance). Mark these indices for removal.

4. Construct the result

Build a new string by including only the characters that are not marked for removal. Return this string as the valid result.

5. Analyze complexity

State the time complexity (O(n)) and space complexity (O(n) for stack or O(1) for counter, excluding output). Discuss potential optimizations.

Key Points to Mention

Stack-based approach: push indices of '(' and pop on matching ')', then remove unmatched indices.
Counter-based approach: track balance and mark invalid ')' when balance is negative, then mark invalid '(' from the end.
Time complexity is O(n) and space complexity can be O(1) with the counter method.
The problem guarantees that a valid result exists after removing minimum parentheses.
Any valid result is acceptable, so multiple solutions are possible.
Edge cases: empty string, string with only parentheses, string with no parentheses.

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