The core idea isn't that hard once you see it: BFS/DFS to mark one island, then BFS outward from it counting steps until you hit the second island.
Use BFS/DFS to identify the two islands and then run a multi-source BFS from one island to find the shortest path to the other, where each step through water costs 1. Alternatively, compute the minimum Manhattan distance between any cell of island A and any cell of island B, but note that this may not always be correct due to obstacles; the BFS approach is more robust.
Pro tip: Clarify that the problem is equivalent to finding the shortest path in a grid where land cells of the same island have cost 0 and water cells have cost 1, and mention that a 0-1 BFS or Dijkstra can be used if movement is allowed in 4 directions. Also, discuss edge cases like islands touching diagonally or the grid being all water except two cells.
Traverse the grid to find the first '1', then use BFS/DFS to mark all cells of that island. Continue scanning to find the second island and mark its cells.
Decide between multi-source BFS from one island or computing pairwise distances. Multi-source BFS is efficient and handles obstacles correctly.
Initialize a queue with all cells of the first island, with distance 0. Expand in 4 directions; when encountering a water cell, increment distance; when encountering a cell of the second island, return the distance.
Use a visited set to avoid reprocessing. Consider if diagonal moves are allowed (usually not). If the islands are already connected, return 0.
Time complexity is O(n^2) since each cell is visited at most once. Space complexity is O(n^2) for the queue and visited set.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.