I knew BFS was involved but my first instinct was to binary search on the answer and check reachability at each candidate value.
Reframe the problem as a decision problem: can we reach the target with all cell values ≤ X? Then binary search on X and use BFS/DFS to check connectivity. Alternatively, use a priority queue (Dijkstra-like) to always expand the cell with the smallest maximum value so far.
Pro tip: Mention that this is a minimax path problem and can be solved optimally with a modified Dijkstra or binary search + BFS; also note that if the grid is small, a simple BFS with a threshold works, but for large grids, the priority queue approach is more efficient.
Confirm the problem: find a path from (0,0) to (n-1,m-1) moving up/down/left/right that minimizes the maximum value along the path. Ask about constraints (grid size, value range) to choose the best algorithm.
Recognize this as a minimax path problem. Two common approaches: (1) binary search on the answer and check feasibility with BFS/DFS, or (2) use a priority queue to always expand the cell with the smallest maximum value so far (Dijkstra-like).
For binary search: set low = max(grid[0][0], grid[n-1][m-1]), high = max value in grid. While low < high, mid = (low+high)/2, check if a path exists using only cells ≤ mid. If yes, high = mid; else low = mid+1. For priority queue: initialize a min-heap with (grid[0][0], 0, 0) and a visited set. While heap not empty, pop the cell with smallest max value; if it's the target, return that value; else push neighbors with max(current_max, neighbor_value).
Binary search + BFS: O(NM log(maxVal)) time, O(NM) space. Priority queue: O(NM log(NM)) time, O(NM) space. Mention that both are efficient for typical grid sizes.
Handle single-cell grid (return its value), grids with all equal values, and unreachable paths (though problem guarantees a path). Compare approaches: binary search is simpler to implement, priority queue may be faster if the answer is small.
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