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

Given a

*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 **invalid signs** on the cells of the `grid`

which points outside the `grid`

.

You will initially start at the upper left cell `(0,0)`

. A valid path in the grid is a path which 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 **doesn't 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:3Explanation: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:0Explanation:You can follow the path from (0, 0) to (2, 2).

**Example 3:**

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

**Example 4:**

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

**Example 5:**

Input:grid = [[4]]Output:0

**Constraints:**

`m == grid.length`

`n == grid[i].length`

`1 <= m, n <= 100`

## Rust Solution

```
struct Solution;
use std::collections::VecDeque;
impl Solution {
fn min_cost(grid: Vec<Vec<i32>>) -> i32 {
let n = grid.len();
let m = grid[0].len();
let mut dist = vec![vec![std::i32::MAX; m]; n];
let mut queue: VecDeque<(usize, usize, i32)> = VecDeque::new();
dist[0][0] = 0;
queue.push_back((0, 0, 0));
while let Some((i, j, d)) = queue.pop_front() {
let right = Self::cost(i, j, d, 1, &grid);
let left = Self::cost(i, j, d, 2, &grid);
let down = Self::cost(i, j, d, 3, &grid);
let up = Self::cost(i, j, d, 4, &grid);
if i > 0 && up < dist[i - 1][j] {
dist[i - 1][j] = up;
queue.push_back((i - 1, j, up));
}
if j > 0 && left < dist[i][j - 1] {
dist[i][j - 1] = left;
queue.push_back((i, j - 1, left));
}
if i + 1 < n && down < dist[i + 1][j] {
dist[i + 1][j] = down;
queue.push_back((i + 1, j, down));
}
if j + 1 < m && right < dist[i][j + 1] {
dist[i][j + 1] = right;
queue.push_back((i, j + 1, right));
}
}
dist[n - 1][m - 1]
}
fn cost(i: usize, j: usize, cost: i32, dir: i32, grid: &[Vec<i32>]) -> i32 {
if dir == grid[i][j] {
cost
} else {
cost + 1
}
}
}
#[test]
fn test() {
let grid = vec_vec_i32![[1, 1, 1, 1], [2, 2, 2, 2], [1, 1, 1, 1], [2, 2, 2, 2]];
let res = 3;
assert_eq!(Solution::min_cost(grid), res);
let grid = vec_vec_i32![[1, 1, 3], [3, 2, 2], [1, 1, 4]];
let res = 0;
assert_eq!(Solution::min_cost(grid), res);
let grid = vec_vec_i32![[2, 2, 2], [2, 2, 2]];
let res = 3;
assert_eq!(Solution::min_cost(grid), res);
let grid = vec_vec_i32![[4]];
let res = 0;
assert_eq!(Solution::min_cost(grid), res);
}
```

Having problems with this solution? Click here to submit an issue on github.