# Rotting Oranges

In a given grid, each cell can have one of three values:

the value 0 representing an empty cell;

the value 1 representing a fresh orange;

the value 2 representing a rotten orange.

Every minute, any fresh orange that is adjacent (4-directionally) to a rotten orange becomes rotten.

Return the minimum number of minutes that must elapse until no cell has a fresh orange. If this is impossible, return -1 instead.

Example 1:

```
Input: [[2,1,1],[1,1,0],[0,1,1]]
Output: 4
```

Example 2:

```
Input: [[2,1,1],[0,1,1],[1,0,1]]
Output: -1
Explanation: The orange in the bottom left corner (row 2, column 0) is never rotten, because rotting only happens 4-directionally.
```

Example 3:

```
Input: [[0,2]]
Output: 0
Explanation: Since there are already no fresh oranges at minute 0, the answer is just 0.
```

Note:

1 <= grid.length <= 10

1 <= grid[0].length <= 10

grid[i][j] is only 0, 1, or 2.

## Solution

```
/**
* @param {number[][]} grid
* @return {number}
* @time complexity: O(m x n)
* @space complexity: O(m x n)
*/
var orangesRotting = function(grid) {
const numRows = grid.length;
const numCols = grid[0].length;
const directions = [
[-1, 0],
[0, 1],
[1, 0],
[0, -1]
];
const queue = [];
let fresh = 0;
let minute = 0;
for (let i = 0; i < numRows; i++) {
for (let j = 0; j < numCols; j++) {
if (grid[i][j] === 2) {
queue.push([i, j]);
} else if (grid[i][j] === 1) {
fresh++;
}
}
}
while (queue.length && fresh) {
const len = queue.length;
for (let q = 0; q < len; q++) {
const rotten = queue.shift();
for (let i = 0; i < directions.length; i++) {
const d = directions[i];
const r = rotten[0] + d[0];
const c = rotten[1] + d[1];
if (r >= 0 && r < numRows && c >= 0 && c < numCols && grid[r][c] === 1) {
grid[r][c] = 2;
fresh--;
queue.push([r, c]);
}
}
}
minute++;
}
if (fresh > 0) {
return -1;
}
return minute;
};
```

## Test Case

```
const assert = require('chai').assert;
describe('Rotting Oranges', () => {
it('should return 4', () => {
assert.strictEqual(orangesRotting([[2,1,1],[1,1,0],[0,1,1]]), 4);
});
it('should return -1 when there is a fresh orange never rotten', () => {
assert.strictEqual(orangesRotting([[2,1,1],[0,1,1],[1,0,1]]), -1);
});
it('should return 0 when there is no fresh oranges', () => {
assert.strictEqual(orangesRotting([[0,2]]), 0);
});
});
```

```
Rotting Oranges
✓ should return 4
✓ should return -1 when there is a fresh orange never rotten
✓ should return 0 when there is no fresh oranges
3 passing (13ms)
```