1631. Path With Minimum Effort

1631. Path With Minimum Effort (Medium)

You are a hiker preparing for an upcoming hike. You are given heights, a 2D array of size rows x columns, where heights[row][col] represents the height of cell (row, col). You are situated in the top-left cell, (0, 0), and you hope to travel to the bottom-right cell, (rows-1, columns-1) (i.e., 0-indexed). You can move up, down, left, or right, and you wish to find a route that requires the minimum effort.

A route's effort is the maximum absolute difference in heights between two consecutive cells of the route.

Return the minimum effort required to travel from the top-left cell to the bottom-right cell.

 

Example 1:

Input: heights = [[1,2,2],[3,8,2],[5,3,5]]
Output: 2
Explanation: The route of [1,3,5,3,5] has a maximum absolute difference of 2 in consecutive cells.
This is better than the route of [1,2,2,2,5], where the maximum absolute difference is 3.

Example 2:

Input: heights = [[1,2,3],[3,8,4],[5,3,5]]
Output: 1
Explanation: The route of [1,2,3,4,5] has a maximum absolute difference of 1 in consecutive cells, which is better than route [1,3,5,3,5].

Example 3:

Input: heights = [[1,2,1,1,1],[1,2,1,2,1],[1,2,1,2,1],[1,2,1,2,1],[1,1,1,2,1]]
Output: 0
Explanation: This route does not require any effort.

 

Constraints:

• rows == heights.length
• columns == heights[i].length
• 1 <= rows, columns <= 100
• 1 <= heights[i][j] <= 106

Related Topics:
Binary Search, Depth-first Search, Union Find, Graph

Similar Questions:

• Path With Maximum Minimum Value (Medium)

Solution 1. Dijkstra

// OJ: https://leetcode.com/problems/path-with-minimum-effort/
// Author: github.com/lzl124631x
// Time: O(MNlog(MN))
// Space: O(MN)
class Solution {
public:
    int minimumEffortPath(vector<vector<int>>& A) {
        int M = A.size(), N = A[0].size(), dirs[4][2] = {{0,1},{0,-1},{1,0},{-1,0}};
        vector<vector<int>> dist(M, vector<int>(N, INT_MAX));
        priority_queue<array<int, 3>, vector<array<int, 3>>, greater<>> pq;
        dist[0][0] = 0;
        pq.push({ 0, 0, 0 });
        while (pq.size()) {
            auto p = pq.top();
            pq.pop();
            int w = p[0], x = p[1], y = p[2];
            if (x == M - 1 && y == N - 1) return dist[x][y];
            if (w > dist[x][y]) continue;
            for (auto &[dx, dy] : dirs) {
                int a = x + dx, b = y + dy;
                if (a < 0 || b < 0 || a >= M || b >= N) continue;
                int d = abs(A[x][y] - A[a][b]);
                if (dist[a][b] > max(dist[x][y], d)) {
                    dist[a][b] = max(dist[x][y], d);
                    pq.push({ dist[a][b], a, b });
                }
            }
        }
        return -1;
    }
};