Databricks·Software Engineer·Technical Phone Screen·Intermediate

Intermediate

Apr 2026

Summary

Databricks SWE interview with a grid pathfinding problem that sounds straightforward until you realize they want you to reason through multiple algorithmic approaches and justify your choices. Felt more like a design discussion than a pure coding round.

Questions Asked (1)

Given an m x n grid where each cell has a non-negative cost, a source cell, and a destination cell, find the minimum cost path from source to destination. You can move to any of the four adjacent cells. The path cost is the sum of all cell costs visited, including source and destination.

Algorithms & Data StructuresTechnical Trade-offs

Author's notes

I started with BFS and they immediately asked what happens when costs aren't uniform.

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AI HintsAI Generated

Suggested Approach

Model the grid as a weighted graph where each cell is a node with cost equal to its value, and edges connect adjacent cells. Use Dijkstra's algorithm with a priority queue to find the minimum cost path from source to destination, treating the cost to enter a cell as the sum of costs along the path. Discuss the time and space complexity and potential optimizations like early termination when the destination is reached.

Pro tip: Mention that if all costs are equal, BFS suffices, but since costs are non-negative and vary, Dijkstra is necessary. Also, highlight that you can avoid modifying the input grid by using a separate distance array.

1. Clarify the problem

Confirm that the path cost includes both source and destination, and that movements are only up, down, left, right. Ask about constraints (e.g., grid size, cost range) to guide algorithm choice.

2. Choose the algorithm

Since edge weights are non-negative (cell costs), Dijkstra's algorithm is optimal. Explain why BFS or DFS would be incorrect or inefficient here.

3. Design the solution

Use a min-heap to store (cost, row, col) and a 2D array to track the minimum cost to reach each cell. Initialize with source cost, then relax neighbors by adding their cell cost.

4. Analyze complexity

Time complexity is O(mn log(mn)) due to heap operations, and space complexity is O(mn) for the distance array and heap. Mention that this is efficient for typical grid sizes.

5. Discuss trade-offs and optimizations

Consider early termination when destination is popped, using a visited set to avoid reprocessing, and potential A* with a heuristic if applicable. Also, note that if costs are uniform, BFS is simpler and faster.

Key Points to Mention

Dijkstra's algorithm is suitable because edge weights (cell costs) are non-negative.
Use a priority queue (min-heap) to always expand the lowest-cost path.
Maintain a distance array to avoid revisiting cells with higher cost.
The cost to enter a cell is added to the cumulative cost from the source.
Time complexity: O(mn log(mn)) with a binary heap; space complexity: O(mn).
Early termination when the destination is extracted from the heap can save time.
If all cell costs are equal, BFS is a simpler alternative.

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