1631. 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:

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

Rust Solution

struct Solution;

use std::cmp::Reverse;
use std::collections::BinaryHeap;

impl Solution {
    fn minimum_effort_path(heights: Vec<Vec<i32>>) -> i32 {
        let n = heights.len();
        let m = heights[0].len();
        let mut visited: Vec<Vec<bool>> = vec![vec![false; m]; n];
        let mut queue: BinaryHeap<(Reverse<i32>, usize, usize)> = BinaryHeap::new();
        let mut res = 0;
        queue.push((Reverse(0), 0, 0));
        while let Some((Reverse(effort), i, j)) = queue.pop() {
            res = res.max(effort);
            if i == n - 1 && j == m - 1 {
                break;
            }
            visited[i][j] = true;
            if i > 0 && !visited[i - 1][j] {
                let diff = heights[i][j] - heights[i - 1][j];
                queue.push((Reverse(diff.abs()), i - 1, j));
            }
            if j > 0 && !visited[i][j - 1] {
                let diff = heights[i][j] - heights[i][j - 1];
                queue.push((Reverse(diff.abs()), i, j - 1));
            }
            if i + 1 < n && !visited[i + 1][j] {
                let diff = heights[i][j] - heights[i + 1][j];
                queue.push((Reverse(diff.abs()), i + 1, j));
            }
            if j + 1 < m && !visited[i][j + 1] {
                let diff = heights[i][j] - heights[i][j + 1];
                queue.push((Reverse(diff.abs()), i, j + 1));
            }
        }
        res
    }
}

#[test]
fn test() {
    let heights = vec_vec_i32![[1, 2, 2], [3, 8, 2], [5, 3, 5]];
    let res = 2;
    assert_eq!(Solution::minimum_effort_path(heights), res);
    let heights = vec_vec_i32![[1, 2, 3], [3, 8, 4], [5, 3, 5]];
    let res = 1;
    assert_eq!(Solution::minimum_effort_path(heights), res);
    let heights = vec_vec_i32![
        [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]
    ];
    let res = 0;
    assert_eq!(Solution::minimum_effort_path(heights), res);
}

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