Databricks·Software Engineer·Technical Phone Screen·Intermediate

Intermediate

May 2026

Summary

Databricks SWE interview with a grid-based algorithm problem that sounds deceptively clean on the surface but has enough moving parts to trip you up if you're not careful about how you handle multiple transportation modes and their separate BFS traversals.

Questions Asked (1)

You have an m x n grid representing a city, and an array of transportation modes where each mode has its own movement rules, passable cells, step sizes, and costs. For each mode, run a BFS from start to destination, then return the minimum cost across all modes. Implement this and walk through the time complexity in terms of m, n, and the number of modes.

Algorithms & Data StructuresTechnical Trade-offs

Author's notes

The BFS part wasn't the hard bit.

Create a free account to read the full note

AI HintsAI Generated

Suggested Approach

Clarify the movement rules and cost structure for each mode, then outline a BFS per mode that respects the mode's step size and passable cells, accumulating costs. After computing the minimum cost for each mode, return the global minimum. Finally, analyze the time complexity as O(k * m * n) where k is the number of modes, and discuss potential optimizations like early termination or multi-source BFS if applicable.

Pro tip: Mention that BFS is optimal for unweighted steps, but if modes have varying costs per step, consider Dijkstra or 0-1 BFS; also note that precomputing passable cells per mode can avoid redundant checks.

1. Clarify problem constraints and assumptions

Ask about movement rules (e.g., 4-directional vs. 8-directional, step sizes), cost accumulation (per step or per cell), and whether modes can be switched mid-path. Confirm if costs are uniform or variable.

2. Design BFS for a single mode

For each mode, implement BFS from start, using a queue and a visited set to avoid cycles. Respect the mode's step size and only enqueue cells that are passable for that mode. Track cumulative cost to each cell.

3. Compute minimum cost per mode and overall

After BFS for a mode, retrieve the cost at the destination if reachable. Keep a running minimum across all modes. If a mode cannot reach the destination, skip it.

4. Analyze time and space complexity

Time: O(k * m * n) where k is number of modes, since each BFS visits each cell at most once. Space: O(m * n) for visited and queue per mode, or O(k * m * n) if storing all modes simultaneously.

5. Discuss optimizations and trade-offs

Consider early termination if a mode's cost exceeds current minimum, using bidirectional BFS, or precomputing passable grids. If costs vary, discuss Dijkstra vs. BFS.

Key Points to Mention

BFS guarantees shortest path in terms of number of steps when all edges have equal weight.
Time complexity is O(k * m * n) because each BFS runs in O(m * n) and we run it for each of k modes.
Space complexity is O(m * n) per BFS, but can be optimized by reusing visited arrays or processing modes sequentially.
If movement costs are not uniform, BFS may not be optimal; consider Dijkstra's algorithm or 0-1 BFS.
Edge cases: unreachable destination, start equals destination, modes with step size > 1, and obstacles.
Potential optimization: early termination when the current BFS cost exceeds the best found so far.

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