Path With Minimum Effort

Path With Minimum Effort

What is this problem about?

The Path With Minimum Effort problem asks you to find a path from the top-left to the bottom-right of a grid minimizing the maximum absolute difference between consecutive cells along the path. This is the classic bottleneck path problem on a grid. The array, matrix, union find, BFS, DFS, binary search, and heap interview pattern is fully exercised.

Why is this asked in interviews?

Cohesity, Microsoft, Meta, Amazon, TikTok, Google, and Bloomberg ask this because it's the "minimize the maximum" grid path problem — the dual of maximizing the minimum. Multiple efficient solutions exist (Dijkstra, binary search + BFS, Union-Find), demonstrating algorithmic versatility.

Algorithmic pattern used

Modified Dijkstra (min bottleneck). effort[i][j] = minimum possible maximum effort to reach (i,j). Use a min-heap of (effort, row, col). Initialize effort[0][0]=0. For each cell: compute new_effort = max(current_effort, |grid[nr][nc] - grid[r][c]|). If new_effort < effort[nr][nc], update and push. Return effort[m-1][n-1].

Example explanation

Grid:

1  2  2
3  8  2
5  3  5

Path (0,0)→(0,1)→(0,2)→(1,2)→(2,2): efforts [1,0,6,3]. Max=6. Path through (1,1): higher effort. Optimal path: 1→2→3→2→5? effort = max(1,2,3)=3? Let me check: 1→3→5→3→5: max(2,2,2,2)=2. Actually optimal = 2 via the bottom path (efforts: |1-3|=2,|3-5|=2,|5-3|=2,|3-5|=2).

Common mistakes candidates make

• Using sum of differences instead of max of differences.
• Not initializing the starting cell's effort to 0.
• Using a max-heap (needs min-heap for minimum effort).
• Not checking if new effort is better before pushing.

Interview preparation tip

Path With Minimum Effort and Path With Maximum Minimum Value are duals of each other. Both use modified Dijkstra — just with different objective (min-max vs max-min). Binary search + BFS is an alternative: binary search on the effort value, BFS to check if a valid path exists within that effort. Practice both approaches and understand their complexity trade-offs (Dijkstra: O(mn log mn), Binary search + BFS: O(mn log(max_diff))).