Path With Minimum Effort

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:


Solution:

class Solution {
    int[] dx = new int[]{-1, 0, 0, 1};
    int[] dy = new int[]{0, -1, 1, 0};
    
    public int minimumEffortPath(int[][] heights) {
        int m = heights.length, n = heights[0].length;
        PriorityQueue<int[]> queue = new PriorityQueue<int[]>((a, b) -> Integer.compare(a[2], b[2]));
        Integer[][] visited = new Integer[m][n]; 
        int min = Integer.MAX_VALUE;
        queue.offer(new int[]{0, 0, 0});
        visited[0][0] = 0;
        while (!queue.isEmpty()) {
            int[] curr = queue.poll();
            if (curr[0] == m - 1 && curr[1] == n - 1) {
                return curr[2];
                // min = Math.min(min, curr[2]);
            }
            for (int k = 0; k < 4; k ++) {
                int nx = curr[0] + dx[k];
                int ny = curr[1] + dy[k];
                if (nx >= 0 && nx < m && ny >= 0 && ny < n) {
                    int[] next = new int[]{nx, ny, Math.max(curr[2], Math.abs(heights[curr[0]][curr[1]] - heights[nx][ny]))};
                    if (visited[nx][ny] == null || visited[nx][ny] > next[2]) {
                        visited[nx][ny] = next[2];
                        queue.offer(next);
                    }
                }
            }
        }
        return min;
    }
}