Formatted question description: https://leetcode.ca/all/1631.html

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:

Solution 1. Dijkstra

// OJ: https://leetcode.com/problems/path-with-minimum-effort/

// 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 [w, x, y] = pq.top();
            pq.pop();
            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;
    }
};

Java

class Solution {
    public int minimumEffortPath(int[][] heights) {
        int rows = heights.length, columns = heights[0].length;
        boolean[][] visited = new boolean[rows][columns];
        int[][] values = new int[rows][columns];
        int[][] directions = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };
        PriorityQueue<int[]> priorityQueue = new PriorityQueue<int[]>(new Comparator<int[]>() {
            public int compare(int[] array1, int[] array2) {
                if (array1[3] != array2[3])
                    return array1[3] - array2[3];
                else if (array1[0] != array2[0])
                    return array1[0] - array2[0];
                else
                    return array1[1] - array2[1];
            }
        });
        priorityQueue.offer(new int[]{0, 0, heights[0][0], values[0][0]});
        while (!priorityQueue.isEmpty()) {
            int[] cell = priorityQueue.poll();
            int row = cell[0], column = cell[1], height = cell[2], value = cell[3];
            for (int i = 0; i < 4; i++) {
                int[] direction = directions[i];
                int newRow = row + direction[0], newColumn = column + direction[1];
                if (newRow >= 0 && newRow < rows && newColumn >= 0 && newColumn < columns) {
                    if (!visited[newRow][newColumn]) {
                        visited[newRow][newColumn] = true;
                        values[newRow][newColumn] = Math.max(value, Math.abs(heights[newRow][newColumn] - height));
                        priorityQueue.offer(new int[]{newRow, newColumn, heights[newRow][newColumn], values[newRow][newColumn]});
                    } else {
                        int newValue = Math.max(value, Math.abs(heights[newRow][newColumn] - height));
                        if (newValue < values[newRow][newColumn]) {
                            values[newRow][newColumn] = Math.max(value, Math.abs(heights[newRow][newColumn] - height));
                            priorityQueue.offer(new int[]{newRow, newColumn, heights[newRow][newColumn], values[newRow][newColumn]});
                        }
                    }
                }
            }
        }
        return values[rows - 1][columns - 1];
    }
}

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