I went with DFS immediately and it was fine.
Treat the grid as a graph and use DFS or BFS to explore each unvisited land cell, marking all connected land cells as visited. Each time you start a traversal from an unvisited land cell, increment the island count. This approach runs in O(m*n) time and space.
Pro tip: Clarify edge cases upfront (empty grid, all water, all land) and mention that you can optimize space by mutating the grid in-place (e.g., setting visited cells to '0') if allowed, but always ask before modifying input.
Confirm the definition of an island, connectivity (4-directional), and input format (e.g., '1' for land, '0' for water). Ask about edge cases like empty grid or large dimensions.
Decide between DFS (recursive or iterative) and BFS. Discuss trade-offs: DFS is simpler but may cause stack overflow on large grids; BFS uses a queue and avoids recursion depth issues.
Iterate through each cell. When you find an unvisited land cell, increment the island count and launch a traversal to mark all connected land cells as visited.
State that time complexity is O(m*n) because each cell is visited once. Space complexity is O(m*n) in the worst case for the recursion stack or queue.
Walk through a small example (e.g., 3x3 grid) to verify correctness. Mention potential pitfalls like diagonal connections or out-of-bounds checks.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
This is the one I was not ready for at the speed they wanted.
Model the grid with a Union-Find (Disjoint Set Union) data structure to dynamically track connected components as land cells are added. For each new land cell, increment the island count, then union it with any adjacent existing land cells, decrementing the count for each successful union. This yields O(α(n)) amortized time per addition, where α is the inverse Ackermann function.
Pro tip: Mention that Union-Find with path compression and union by rank is the optimal solution, and briefly discuss how you would handle edge cases like duplicate land additions or out-of-bounds coordinates. This shows you consider robustness and real-world constraints.
Ask about grid size, whether coordinates are guaranteed unique, and if the stream can contain duplicate or invalid cells. Confirm that islands are 4-directionally connected.
Select Union-Find (Disjoint Set Union) to efficiently manage connected components. Explain why a simple BFS/DFS per addition would be too slow (O(N) per query).
For each new land cell: increment island count, mark it as land, then check its four neighbors. For each neighbor that is land, union the two cells; if they were in different sets, decrement the island count.
State that each addition takes O(α(N)) amortized time with path compression and union by rank, where N is the number of cells. Space is O(N) for the parent and rank arrays.
Discuss duplicate additions (ignore or handle gracefully), out-of-bounds coordinates, and the initial empty grid. Walk through a small example to verify correctness.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.