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My first instinct was DFS and I coded that up fine, but then they asked me to redo it with BFS and I fumbled a bit with the queue initialization.
Start by clarifying the problem constraints and edge cases, then explain the BFS approach: iterate through each cell, and when a '1' is found, increment the island count and use BFS to mark all connected land cells as visited. Emphasize that BFS uses a queue to explore neighbors level by level, and discuss how to handle edge cases like empty grid, all water, all land, and large grids.
Pro tip: Mention that you can optimize space by marking visited cells in-place (e.g., changing '1' to '0') instead of using a separate visited matrix, and discuss the trade-offs. Also, be prepared to compare BFS with DFS and explain why BFS might be preferred in certain scenarios (e.g., avoiding recursion depth issues).
Ask clarifying questions about grid dimensions, input format, and definition of an island (4-directional vs 8-directional connectivity). State assumptions clearly.
Explain that you'll iterate through each cell; when a '1' is found, increment count and perform BFS to mark all connected '1's as visited. Use a queue to explore neighbors.
Describe the BFS steps: enqueue the starting cell, mark it visited, then while queue not empty, dequeue and check its 4 neighbors; if a neighbor is '1' and unvisited, mark and enqueue. Use directions array for clarity.
Discuss edge cases: empty grid, grid with no land, grid with all land, single row/column, and large grids. Explain how BFS handles them (e.g., check bounds before accessing neighbors).
State time complexity O(M*N) and space complexity O(min(M,N)) for BFS queue in worst case. Mention in-place marking to save space and compare with DFS.
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