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1368. Minimum Cost to Make at Least One Valid Path in a Grid

Given an m x n grid. Each cell of the grid has a sign pointing to the next cell you should visit if you are currently in this cell. The sign of grid[i][j] can be:

  • 1 which means go to the cell to the right. (i.e go from grid[i][j] to grid[i][j + 1])
  • 2 which means go to the cell to the left. (i.e go from grid[i][j] to grid[i][j - 1])
  • 3 which means go to the lower cell. (i.e go from grid[i][j] to grid[i + 1][j])
  • 4 which means go to the upper cell. (i.e go from grid[i][j] to grid[i - 1][j])

Notice that there could be some signs on the cells of the grid that point outside the grid.

You will initially start at the upper left cell (0, 0). A valid path in the grid is a path that starts from the upper left cell (0, 0) and ends at the bottom-right cell (m - 1, n - 1) following the signs on the grid. The valid path does not have to be the shortest.

You can modify the sign on a cell with cost = 1. You can modify the sign on a cell one time only.

Return the minimum cost to make the grid have at least one valid path.

Example 1:

Input: grid = [[1,1,1,1],[2,2,2,2],[1,1,1,1],[2,2,2,2]]
Output: 3
Explanation: You will start at point (0, 0).
The path to (3, 3) is as follows. (0, 0) --> (0, 1) --> (0, 2) --> (0, 3) change the arrow to down with cost = 1 --> (1, 3) --> (1, 2) --> (1, 1) --> (1, 0) change the arrow to down with cost = 1 --> (2, 0) --> (2, 1) --> (2, 2) --> (2, 3) change the arrow to down with cost = 1 --> (3, 3)
The total cost = 3.

Example 2:

Input: grid = [[1,1,3],[3,2,2],[1,1,4]]
Output: 0
Explanation: You can follow the path from (0, 0) to (2, 2).

Example 3:

Input: grid = [[1,2],[4,3]]
Output: 1

Constraints:

  • m == grid.length
  • n == grid[i].length
  • 1 <= m, n <= 100
  • 1 <= grid[i][j] <= 4

Solutions (Rust)

1. Solution

impl Solution {
    pub fn min_cost(grid: Vec<Vec<i32>>) -> i32 {
        use std::collections::BinaryHeap;
        use std::collections::HashSet;

        let m = grid.len();
        let n = grid[0].len();
        let mut seen = HashSet::new();
        let mut heap = BinaryHeap::from([(0, 0, 0)]);

        while let Some((cost, i0, j0)) = heap.pop() {
            if i0 < 0 || j0 < 0 || i0 >= m || j0 >= n {
                continue;
            }

            if i0 == m - 1 && j0 == n - 1 {
                return -cost;
            }

            seen.insert((i0, j0));

            for (i1, j1, g) in [
                (i0, j0 + 1, 1),
                (i0, j0 - 1, 2),
                (i0 + 1, j0, 3),
                (i0 - 1, j0, 4),
            ] {
                if !seen.contains(&(i1, j1)) {
                    if g == grid[i0 as usize][j0 as usize] {
                        heap.push((cost, i1, j1));
                    } else {
                        heap.push((cost - 1, i1, j1));
                    }
                }
            }
        }

        unreachable!()
    }
}