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Bloomberg·Software Engineer·Technical Phone Screen·Intermediate

Intermediate
Jun 2026

Summary

Bloomberg coding screen for a software engineer role, one question, grid-based BFS problem that's basically a reskin of a well-known leetcode problem. Pretty straightforward if you've seen it before.

Questions Asked (1)

Q1

Given a 2D grid where cells can be empty, inflated, or popped balloons, a popped balloon causes adjacent inflated balloons to also pop. Determine whether all inflated balloons can eventually be popped.

Algorithms & Data Structures
Author's notes

Recognized it pretty fast as the rotting oranges problem with a costume on.

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AI HintsAI Generated

Suggested Approach

Model the grid as a graph where each inflated balloon is a node and edges connect adjacent inflated balloons. The problem reduces to checking if the graph is connected; if so, popping any balloon will eventually pop all, otherwise some remain. Use BFS/DFS to traverse from any inflated balloon and count reachable nodes.

Pro tip: Clarify with the interviewer whether popping is simultaneous or sequential, as this affects the algorithm. Also, mention that if the grid is large, an in-place BFS can save memory.

1. Clarify the problem

Ask clarifying questions: Are diagonal adjacencies considered? Does popping propagate only to directly adjacent balloons or also to those popped as a result? Is the grid static or can we choose the order of popping?

2. Model as a graph

Treat each inflated balloon as a vertex. Add edges between adjacent inflated balloons (up, down, left, right). The problem becomes: is this graph connected?

3. Choose traversal algorithm

Use BFS or DFS to traverse the graph starting from any inflated balloon. Count the number of visited inflated balloons.

4. Compare counts

If the number of visited inflated balloons equals the total number of inflated balloons, then all can be popped; otherwise, they cannot.

5. Analyze complexity and edge cases

Discuss time and space complexity (O(R*C) time, O(R*C) space for visited). Handle edge cases: no inflated balloons, all inflated, disconnected components.

Key Points to Mention

  • Graph connectivity: the problem is equivalent to checking if the inflated balloons form a single connected component.
  • Adjacency definition: typically 4-directional (up, down, left, right), but confirm with interviewer.
  • Traversal algorithm: BFS or DFS can be used; BFS is often preferred for shortest path but both work.
  • Time and space complexity: O(R*C) time and O(R*C) space for visited set or queue.
  • Edge cases: empty grid, no inflated balloons, all inflated balloons, multiple disconnected components.
  • Optimization: in-place marking (e.g., changing 'inflated' to 'visited') to save space if allowed.

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