Clarify the problem constraints and edge cases, then propose a DFS or BFS solution that iterates through each cell, and when a '1' is found, increments the island count and sinks the entire island by marking connected land cells as '0'. Analyze time and space complexity, and discuss potential optimizations or trade-offs.
Pro tip: Mention that you can avoid modifying the input by using a separate visited set, but if modification is allowed, in-place marking is more space-efficient. Also, discuss how to handle very large grids that don't fit in memory, showing awareness of scalability.
Ask about grid size, whether the input can be modified, and if diagonal connections count. Confirm that the grid is rectangular and contains only '0's and '1's.
Decide between DFS (recursive or iterative) and BFS. Mention that DFS is simpler but may cause stack overflow for large grids; BFS uses a queue and is safer for deep recursion.
Iterate through each cell. When a '1' is found, increment island count and perform DFS/BFS to mark all connected land cells as visited (e.g., set to '0'). Continue until all cells are processed.
Time complexity is O(M×N) since each cell is visited once. Space complexity is O(M×N) in the worst case for the recursion stack or queue, but can be O(min(M,N)) with optimized BFS.
Handle empty grid, all water, all land, and single row/column. Mention iterative DFS to avoid recursion limits, and union-find as an alternative approach.
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