Classic BFS/DFS problem and I knew it, but I still fumbled explaining the visited tracking out loud.
Treat the grid as a graph where each land cell is a node connected to its adjacent land cells. Use DFS or BFS to explore and mark all cells in each connected component, incrementing the island count for each unvisited land cell encountered.
Pro tip: Clarify edge cases upfront (empty grid, all water, all land) and mention that you can mutate the grid to mark visited cells to save space, but ask if that's acceptable or if you should preserve the input.
Ask about grid dimensions, connectivity (4-directional vs 8-directional), and whether the grid can be modified. Confirm input/output format and edge cases.
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 a '1' is found, increment island count and launch a traversal 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 the worst case for the recursion stack or queue.
Walk through a small example, including edge cases like empty grid, single row/column, and all land or all water, to verify correctness.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Use DFS or BFS to identify each island, then normalize its shape by translating all coordinates relative to the top-leftmost cell. Store the normalized shape in a hash set to count distinct shapes.
Pro tip: Mention that you can optimize space by using a canonical string representation of the normalized coordinates, and discuss trade-offs between DFS and BFS for large grids.
Confirm that islands are connected components of 1s (4-directionally) and that translation means shifting without rotation or reflection. Ask about grid size and constraints.
Select DFS (recursive or iterative) or BFS to explore each island. Consider recursion depth for large grids and potential stack overflow.
During traversal, record the coordinates of each cell. After traversal, translate all coordinates so that the minimum row and column are 0, making the shape invariant to translation.
Convert the normalized coordinates to a hashable form (e.g., a string or tuple of sorted coordinates) and insert into a set. The set size is the number of distinct shapes.
Time complexity is O(R*C) for traversal and normalization. Space complexity is O(R*C) for the visited set and shape storage. Discuss potential optimizations.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.