My first instinct was just BFS from one island toward the other, which is mostly right, but I fumbled the setup.
Model the problem as a shortest path between two sets of cells: first identify the two islands via BFS/DFS, then run a multi-source BFS from one island to find the minimum distance to the other, where distance is the number of water cells crossed. Alternatively, compute the minimum Manhattan distance between any land cell of island A and any land cell of island B, but be careful because the shortest path may not be a straight line due to obstacles.
Pro tip: Clarify whether flipping a water cell to land can connect diagonally or only 4-directionally; the problem states 4-directionally connected islands, so the connection must also be 4-directionally. Also, mention that the answer is at least 1 because the islands are separate, and you can use 0-1 BFS or multi-source BFS to handle the cost of flipping water cells efficiently.
Use BFS or DFS to label each land cell with its island ID (1 or 2). This separates the grid into two sets of coordinates.
Decide between multi-source BFS from one island to the other, or computing minimum Manhattan distance between all pairs of land cells. Multi-source BFS is more robust for obstacles.
Initialize a queue with all cells of island 1, and BFS layer by layer. Each step into a water cell increments the distance by 1. Stop when you reach any cell of island 2.
The BFS distance when first hitting island 2 is the minimum number of water cells to flip. If using Manhattan distance, compute min over all pairs of (|r1-r2| + |c1-c2| - 1) but verify with BFS.
Time O(R*C), space O(R*C). Discuss edge cases: islands adjacent diagonally (answer 1), large grid, and ensuring BFS doesn't revisit cells.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.