Google·Software Engineer·Technical Phone Screen·Intermediate

IntermediatePrefer not to say

Jun 2026

Summary

Google SWE interview with a classic grid traversal problem. Nothing too exotic but they wanted you to actually think through multiple approaches rather than just code the first thing that popped into your head.

Questions Asked (1)

Given a 2D binary grid of '1's (land) and '0's (water), count the number of islands, where an island is a group of adjacent land cells connected horizontally or vertically. Walk through multiple approaches including depth-first search, breadth-first search, and union-find.

Algorithms & Data StructuresTechnical Trade-offs

Author's notes

I went straight to DFS because it's the one I can code fastest under pressure, which maybe wasn't the smartest move.

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AI HintsAI Generated

Suggested Approach

Start by clarifying the problem and constraints, then present a clear traversal-based solution (DFS or BFS) with O(m*n) time and space complexity. Follow up by discussing alternative approaches like Union-Find, comparing their trade-offs in terms of performance, code complexity, and suitability for different scenarios.

Pro tip: Mention that you can optimize space by marking visited cells in-place (e.g., changing '1' to '0') to avoid a separate visited matrix, and note that Union-Find can be more efficient for dynamic or streaming inputs.

1. Clarify and Define

Restate the problem to ensure understanding: count connected components of '1's in a 2D grid using 4-directional adjacency. Discuss edge cases like empty grid, all water, or all land.

2. Traversal Approach (DFS/BFS)

Explain that you can iterate through each cell; when you find a '1', increment the island count and use DFS or BFS to mark all connected land cells as visited. Highlight that both have O(m*n) time and space complexity.

3. Union-Find Approach

Describe how to use Union-Find (Disjoint Set Union) to group adjacent land cells. Initialize each land cell as a separate set, union with adjacent land cells, and count distinct roots. Mention path compression and union by rank for efficiency.

4. Compare Trade-offs

Compare DFS, BFS, and Union-Find: DFS/BFS are simpler and use less memory for sparse grids; Union-Find is better for dynamic connectivity or when the grid is large and recursion depth is a concern. Discuss iterative vs recursive DFS to avoid stack overflow.

5. Complexity and Optimization

State time and space complexities for each approach. Mention in-place modification to save space and potential optimizations like early termination or using a queue for BFS.

Key Points to Mention

Time and space complexity: O(m*n) for all approaches, with space O(m*n) for visited or Union-Find structures.
In-place modification: changing '1' to '0' to mark visited, reducing space to O(1) extra for DFS/BFS (excluding recursion stack).
Recursive DFS may cause stack overflow for large grids; iterative DFS or BFS is safer.
Union-Find with path compression and union by rank achieves near O(1) amortized per operation.
Handling edge cases: empty grid, single row/column, all water, all land.
Choice of approach depends on context: DFS/BFS for static grids, Union-Find for dynamic or streaming scenarios.

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