[Leetcode] Problem 1631 - Path With Minimum Effort

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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

No.1

B8G4z9.png

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.

No.2

B8GOiD.png

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].

No.3

B8GXJe.png

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] <= 10^6

Code

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public int minimumEffortPath(int[][] heights) {
    int m = heights.length;
    int n = heights[0].length;
    int[][] dirs = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
    int[] efforts = new int[m * n];
    Arrays.fill(efforts, (int) 1e6 + 1);
    PriorityQueue<int[]> pq = new PriorityQueue<>((a, b) -> a[0] - b[0]);
    pq.offer(new int[] {0, 0, 0});

    while (!pq.isEmpty()) {
        int[] node = pq.poll();

        if (node[1] == m - 1 && node[2] == n - 1)
            return node[0];

        for (int[] dir : dirs) {
            int x = node[1] + dir[0];
            int y = node[2] + dir[1];

            if (x < 0 || y < 0 || x >= m || y >= n)
                continue;

            int effort = Math.max(node[0], Math.abs(heights[x][y] - heights[node[1]][node[2]]));

            if (effort < efforts[x * n + y]) {
                efforts[x * n + y] = effort;
                pq.offer(new int[] {effort, x, y});
            }
        }
    }

    return -1;
}