My first instinct was to just do a BFS and track the path sum, which was completely wrong.
Model the problem as finding a path that minimizes the maximum cell value, which can be solved using binary search on the answer combined with BFS/DFS, or by using a priority queue (Dijkstra-like) to always expand the cell with the smallest maximum height so far. Start from the top-left, and at each step, choose the neighbor that minimizes the maximum height encountered. The first time you reach the bottom-right, the current maximum is the answer.
Pro tip: Clarify edge cases upfront (e.g., single cell, unreachable destination) and mention that the optimal solution runs in O(N log N) with a priority queue, but binary search + BFS is O(N log(maxHeight)) which is also acceptable. This shows you consider trade-offs.
Restate the problem: find a path from (0,0) to (n-1,m-1) minimizing the maximum cell value along the path. Clarify that you can move in four directions and that the water level must be at least the cell's height.
Consider two main approaches: (1) binary search on the answer and check feasibility with BFS/DFS, or (2) use a priority queue (Dijkstra-like) to always expand the cell with the smallest maximum height so far. Explain why both work and their complexities.
For binary search: define low and high as min and max cell values, then for each mid, run BFS from start to end only traversing cells with height <= mid. For priority queue: use a min-heap storing (maxHeightSoFar, row, col), and update the max when moving to a neighbor.
Discuss time and space complexity: binary search O(N log(maxHeight)) with BFS O(N), priority queue O(N log N). Handle edge cases: single cell, no path, all cells same height.
Walk through a small example to verify correctness. Summarize the solution and mention potential optimizations or alternative approaches.
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