The core idea is BFS in two phases: first find one island completely, then do a multi-source BFS expanding outward until you hit the second island.
Model the problem as finding the shortest path between the two islands where traversing water costs 1 and land costs 0. Use multi-source BFS from one island, expanding through water cells, and return the minimum distance to reach the other island.
Pro tip: Clarify that flipping a water cell to land connects it, so the answer is the minimum number of water cells on a path between the islands. Mention that 0-1 BFS or Dijkstra can also be used, but multi-source BFS is optimal for this binary cost scenario.
Use DFS or BFS to find all cells of the first island and mark them. The second island is the remaining unvisited land cells.
Add all cells of the first island to a queue with distance 0. These are the starting points for expansion.
Perform BFS where moving to a water cell costs 1 and to a land cell costs 0. Use a deque for 0-1 BFS or a priority queue for Dijkstra to handle varying costs.
When a cell belonging to the second island is reached, return the accumulated cost as the minimum number of water cells to flip.
Consider grids where islands are adjacent (answer 0) or where multiple shortest paths exist. Ensure the algorithm terminates correctly.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.