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Bloomberg·Software Engineer·Technical Phone Screen·Intermediate

Intermediate
May 2026

Summary

Bloomberg SWE interview with a classic but tricky parentheses problem. The question had some depth to it once you got past the surface reading.

Questions Asked (1)

Q1

Given a string with parentheses and other characters, remove the fewest parentheses possible to make it valid. Return all unique valid results.

Algorithms & Data StructuresTechnical Trade-offs
Author's notes

I went with BFS first because it felt safer to explain level by level, but the interviewer pushed me on efficiency pretty quickly.

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AI HintsAI Generated

Suggested Approach

Use BFS level-by-level to find the minimum number of removals, checking each string for validity and stopping at the first level where valid strings appear. At each level, generate all unique strings by removing one parenthesis and use a set to avoid duplicates and repeated work. Collect all valid strings at that level and return them.

Pro tip: Emphasize that BFS guarantees the minimum removals and that using a visited set prevents exponential blowup; also mention that you can precompute the minimum removals needed with a single pass to potentially prune the search.

1. Clarify and Define Validity

Confirm that a valid string has balanced parentheses and that other characters are ignored. Define what constitutes a unique result and how to handle empty strings.

2. Choose BFS for Minimum Removals

Explain that BFS explores all strings with k removals before k+1, ensuring the first valid strings found use the fewest removals. Use a queue and a visited set to manage states.

3. Generate and Validate

At each step, pop a string, check if it's valid; if so, add to results. If not, generate all strings by removing one parenthesis, and enqueue those not seen before.

4. Collect and Return Results

Once valid strings are found at a level, continue processing that level to collect all unique valid strings, then return them without processing further levels.

5. Analyze Complexity and Trade-offs

Discuss time and space complexity, noting that BFS can be exponential in worst case but is optimal for minimum removals. Mention alternative approaches like DFS with pruning or DP.

Key Points to Mention

  • BFS ensures minimum removals by exploring level by level.
  • Use a set to track visited strings and avoid duplicate work.
  • Validity check: balance counter never negative and ends at zero.
  • Only remove parentheses, not other characters.
  • Stop at the first level where valid strings are found.
  • Worst-case time complexity is O(2^n) but pruning helps.

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