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.
1 class Solution { 2 public int minimumEffortPath(int[][] heights) { 3 int[] DIR = new int[] { 0, 1, 0, -1, 0 }; 4 int m = heights.length, n = heights[0].length; 5 int[][] dist = new int[m][n]; 6 for (int i = 0; i < m; i++) { 7 Arrays.fill(dist[i], Integer.MAX_VALUE); 8 } 9 10 Queue<int[]> minHeap = new PriorityQueue<>(Comparator.comparingInt(a -> a[0])); 11 minHeap.offer(new int[] { 0, 0, 0 }); // distance, row, col 12 while (!minHeap.isEmpty()) { 13 int[] top = minHeap.poll(); 14 int d = top[0], r = top[1], c = top[2]; 15 if (r == m - 1 && c == n - 1) return d; // Reach to bottom right 16 for (int i = 0; i < 4; i++) { 17 int nr = r + DIR[i], nc = c + DIR[i + 1]; 18 if (nr >= 0 && nr < m && nc >= 0 && nc < n) { 19 int newDist = Math.max(d, Math.abs(heights[nr][nc] - heights[r][c])); 20 if (dist[nr][nc] > newDist) { 21 dist[nr][nc] = newDist; 22 minHeap.offer(new int[] { dist[nr][nc], nr, nc }); 23 } 24 } 25 } 26 } 27 return 0; // Unreachable code, Java require to return interger value. 28 } 29 }