README.md

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

Solutions (Rust)

1. Solution

impl Solution {
    pub fn minimum_effort_path(heights: Vec<Vec<i32>>) -> i32 {
        let mut min_efforts = vec![vec![i32::MAX; heights[0].len()]; heights.len()];
        let mut cells = vec![(0, 0)];
        min_efforts[0][0] = 0;

        while let Some((i, j)) = cells.pop() {
            if i > 0 {
                let effort = (heights[i][j] - heights[i - 1][j])
                    .abs()
                    .max(min_efforts[i][j]);

                if effort < min_efforts[i - 1][j] {
                    min_efforts[i - 1][j] = effort;
                    cells.push((i - 1, j));
                }
            }
            if i < heights.len() - 1 {
                let effort = (heights[i][j] - heights[i + 1][j])
                    .abs()
                    .max(min_efforts[i][j]);

                if effort < min_efforts[i + 1][j] {
                    min_efforts[i + 1][j] = effort;
                    cells.push((i + 1, j));
                }
            }
            if j > 0 {
                let effort = (heights[i][j] - heights[i][j - 1])
                    .abs()
                    .max(min_efforts[i][j]);

                if effort < min_efforts[i][j - 1] {
                    min_efforts[i][j - 1] = effort;
                    cells.push((i, j - 1));
                }
            }
            if j < heights[0].len() - 1 {
                let effort = (heights[i][j] - heights[i][j + 1])
                    .abs()
                    .max(min_efforts[i][j]);

                if effort < min_efforts[i][j + 1] {
                    min_efforts[i][j + 1] = effort;
                    cells.push((i, j + 1));
                }
            }
        }

        *min_efforts.last().unwrap().last().unwrap()
    }
}