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Classic BFS/DFS problem and I still fumbled the edge case handling for a second.
Treat the grid as a graph and use DFS/BFS to explore each island, marking visited cells to avoid recounting. Iterate through each cell; when you find an unvisited '1', increment the island count and flood-fill to mark all connected land.
Pro tip: Clarify edge cases upfront (empty grid, all water, all land) and mention that you can optimize space by modifying the grid in-place if allowed, or use a separate visited matrix if not.
Confirm grid dimensions, connectivity definition (4-directional), and whether input can be modified. Discuss edge cases like empty grid or no land.
Decide between DFS (recursive or iterative) and BFS. Consider trade-offs: DFS is simpler but may risk stack overflow on large grids; BFS uses a queue and avoids recursion depth issues.
Write a helper function that, given a starting cell, marks all connected land cells as visited. Use a visited set/matrix or modify the grid in-place (e.g., change '1' to '0').
Iterate through every cell in the grid. When encountering an unvisited '1', increment the island count and invoke the traversal to mark the entire island.
State time complexity O(M×N) since each cell is visited once, and space complexity O(M×N) in worst case for recursion/queue or visited matrix.
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