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Amazon·Software Engineer·Technical Phone Screen·Intermediate

Intermediate
May 2026

Summary

Amazon SWE coding round, one problem the whole session. Grid traversal, which sounds straightforward until you actually have to think about edge cases under pressure with someone watching you.

Questions Asked (1)

Q1

Given a 2D grid of elevations and a starting cell, find all cells that water can reach by flowing only to strictly lower neighbors in 4 directions.

Algorithms & Data StructuresTechnical Trade-offs
Author's notes

BFS or DFS from the start, only expanding to neighbors with strictly lower height.

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AI HintsAI Generated

Suggested Approach

Clarify the problem constraints and edge cases, then propose a reverse BFS/DFS from the starting cell to find all reachable cells. Discuss time and space complexity, and compare with alternative approaches like forward DFS with memoization.

Pro tip: Mention that reversing the flow direction simplifies the traversal because you can start from the given cell and explore neighbors with higher or equal elevation, avoiding the need to track visited cells in multiple directions. This also naturally handles cycles and ensures each cell is visited once.

1. Clarify the problem

Ask about grid size, elevation range, whether diagonal moves are allowed, and if the starting cell is included in the result. Confirm that water flows only to strictly lower neighbors.

2. Choose an approach

Propose a reverse BFS/DFS starting from the given cell, exploring neighbors with elevation >= current cell's elevation. Alternatively, consider forward DFS with memoization, but note the reverse approach is simpler and more efficient.

3. Walk through the algorithm

Initialize a queue with the starting cell, mark it visited, and while the queue is not empty, pop a cell and check its 4 neighbors. If a neighbor is unvisited and its elevation is >= the current cell's elevation, add it to the queue and mark visited.

4. Analyze complexity

State that time complexity is O(m*n) where m and n are grid dimensions, as each cell is visited at most once. Space complexity is O(m*n) for the visited set and queue in the worst case.

5. Discuss edge cases and trade-offs

Mention handling of empty grid, single cell, all equal elevations, and disconnected regions. Compare BFS vs DFS (both work, BFS avoids recursion depth issues) and note that the reverse approach avoids redundant checks.

Key Points to Mention

  • Reverse BFS/DFS from the starting cell to find all cells that can flow to it.
  • Condition for traversal: neighbor's elevation >= current cell's elevation (since water flows from higher to lower).
  • Use a visited set to avoid revisiting cells and handle cycles.
  • Time and space complexity: O(m*n) for both, where m*n is the number of cells.
  • Edge cases: empty grid, single cell, all elevations equal, starting cell at boundary.
  • Comparison with forward DFS: reverse approach is more efficient and simpler to implement.

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