This one is tricky if you haven't seen it before.
Model the grid as a graph where each cell's elevation is the cost to enter it, and the goal is to find a path from (0,0) to (n-1,n-1) that minimizes the maximum elevation along the path. Use either a modified Dijkstra's algorithm with a priority queue or binary search on the answer combined with BFS/DFS to check feasibility.
Pro tip: Clarify that the problem is equivalent to finding the minimax path, and mention that both Dijkstra and binary search+BFS have the same time complexity O(n^2 log n), but Dijkstra is often simpler to implement and explain.
Restate the problem: find the minimum time t such that there is a path from top-left to bottom-right where all cells have elevation ≤ t. Note that you can wait, but waiting doesn't help because you can always start later.
Decide between Dijkstra-like search (minimax path) or binary search on t with BFS/DFS feasibility check. Both are valid; pick the one you can implement most cleanly under pressure.
For Dijkstra: use a min-heap storing (max_elevation_so_far, row, col), and update when a smaller max elevation is found. For binary search: define check(t) that runs BFS from start to end using only cells with elevation ≤ t.
Time complexity: O(n^2 log n) for both approaches. Space: O(n^2). Handle edge cases: n=1 (return grid[0][0]), and ensure you don't revisit cells unnecessarily.
Walk through a small example to verify correctness. Mention that you can optimize binary search bounds to [max(grid[0][0], grid[n-1][n-1]), max(grid)] or use a union-find approach if asked for alternatives.
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