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I went with DFS right away which was fine, but I fumbled a bit explaining the space complexity.
Clarify the problem constraints (e.g., grid size, definition of connectivity) and then propose a graph traversal solution (DFS/BFS) to explore each island. Discuss trade-offs between recursive and iterative approaches, and mention potential optimizations like union-find or in-place marking.
Pro tip: Demonstrate awareness of real-world implications: for large grids, recursion may cause stack overflow, so an iterative BFS or union-find with path compression is more robust. Also, consider if the grid can be modified in-place to save space.
Ask about edge cases: empty grid, non-binary values, connectivity definition (4-directional vs 8-directional), and whether the grid can be mutated. Confirm input/output format.
Explain that you'll iterate through each cell; when you find a '1', increment the island count and use DFS/BFS to mark all connected '1's as visited (e.g., set to '0' or use a visited set).
State time complexity O(M*N) since each cell is visited once, and space complexity O(M*N) in worst case for recursion stack or queue, or O(1) if modifying in-place with iterative DFS.
Compare DFS (simple, but recursion depth risk) vs BFS (iterative, uses queue) vs Union-Find (good for dynamic connectivity, but more complex). Mention that in-place modification saves space but may not be allowed.
Walk through a small example, test edge cases like all water, all land, and single row/column. Mention potential pitfalls like stack overflow for large grids and how to mitigate.
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