My first instinct was pure BFS from every water cell and I wasted probably five minutes going down that path before realizing I needed to find each island first.
Model the problem as a shortest path on a graph where each cell is a node and edges connect adjacent cells with weight 0 for land and 1 for water. Use multi-source 0-1 BFS starting from all cells of one island to find the minimum cost to reach the other island, where cost equals the number of water cells flipped. Alternatively, label the islands and compute the minimum Manhattan distance between their cells minus 1.
Pro tip: Clarify assumptions upfront: whether flipping a 0 to 1 can connect diagonally or only orthogonally, and whether the two islands are guaranteed to exist. Mentioning edge cases and constraints (e.g., grid size) shows thoroughness and can guide the interviewer's expectations.
Confirm the problem details: connectivity definition (4-directional vs 8-directional), input guarantees (exactly two islands), and output (minimum flips). Restate the problem in your own words to ensure alignment.
Use BFS/DFS to find and label the two separate islands. Store the coordinates of cells belonging to each island for later use.
Treat each cell as a node; edges between adjacent cells have weight 0 if both are land, weight 1 if one is water (representing a flip). The goal is to find the minimum total weight path from any cell of island A to any cell of island B.
Initialize a deque with all cells of island A at distance 0. Perform 0-1 BFS: when moving to a land cell, push front; to a water cell, push back and increment distance. Stop when reaching any cell of island B; the distance is the minimum flips.
State time and space complexity: O(R*C) time and space. Discuss potential optimizations like early termination or using Manhattan distance if only orthogonal moves are allowed and no obstacles exist.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.