Openai·Software Engineer·Technical Phone Screen·Intermediate

Intermediate

Apr 2026

Summary

Coding round for a software engineer role at OpenAI. Single algorithmic problem, grid-based BFS. Pretty standard for this kind of company but the multi-source angle is where people slip up.

Questions Asked (1)

You're given an m x n grid where each cell is either empty, a fresh orange, or a rotten orange. Each minute, a fresh orange adjacent (in 4 directions) to a rotten one also becomes rotten. What's the minimum number of minutes until no fresh oranges remain? Return -1 if it's not possible.

Algorithms & Data Structures

Author's notes

The BFS part clicked pretty fast but I almost started with a single-source approach and would've gotten the wrong answer.

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AI HintsAI Generated

Suggested Approach

Model the problem as a multi-source BFS where all initially rotten oranges are enqueued at time 0. Process level by level, incrementing time each level, and track the number of fresh oranges to detect impossibility. Return the final time if all fresh oranges rot, else -1.

Pro tip: Clarify edge cases upfront (e.g., no fresh oranges, no rotten oranges) and mention that BFS is optimal because each orange rots at the earliest possible minute. Also, discuss space-time trade-offs and potential optimizations like in-place marking.

1. Understand the problem and edge cases

Restate the problem to ensure clarity: grid with fresh (1), rotten (2), empty (0). Identify edge cases: no fresh oranges (return 0), no rotten oranges but fresh present (return -1), all empty (return 0).

2. Choose BFS as the optimal approach

Explain why BFS is ideal: it simulates the simultaneous rotting process level by level, ensuring minimum time. DFS or simulation would be inefficient.

3. Initialize the BFS queue and count fresh oranges

Scan the grid to enqueue all rotten oranges and count fresh oranges. This sets up the multi-source BFS starting point.

4. Perform level-order traversal

Process the queue level by level, incrementing time after each level. For each rotten orange, check its 4 neighbors; if fresh, mark rotten, decrement fresh count, and enqueue.

5. Return the result

After BFS, if fresh count is 0, return the elapsed time; otherwise, return -1. Discuss time and space complexity: O(m*n) time and space.

Key Points to Mention

Multi-source BFS to handle simultaneous rotting from all initially rotten oranges.
Tracking the number of fresh oranges to determine if all can be rotted.
Level-by-level processing to correctly compute minutes.
Edge cases: no fresh oranges, no rotten oranges, all empty, unreachable fresh oranges.
Time and space complexity: O(m*n) for both, as each cell is processed once.
Potential optimizations: in-place modification of the grid to avoid extra space, though queue still needed.

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