leetcode/apps/docs/content/(04-relational)/graph-bfs/994-rotting-oranges.mdx

---
title: '994. Rotting Oranges'
description: 'You are given an m x n grid where each cell can have one of three values:'
icon: dot
sidebar:
  label: 'Rotting Oranges'
  badge: 'Medium'
---

<Badge variant="accent">Array</Badge>

### Example 1:
- Input: `grid = [[2,1,1],[1,1,0],[0,1,1]]`
- Output: `4`

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

### Constraints:

- `m == grid.length`
- `n == grid[i].length`
- `1 <= m, n <= 10`
- grid[i][j] is 0, 1, or 2.

## Solution

```py
from collections import deque


class Solution:
    def orangesRotting(self, grid: list[list[int]]) -> int:
        q = deque()
        time, fresh = 0, 0

        rows, cols = len(grid), len(grid[0])
        for r in range(rows):
            for c in range(cols):
                if grid[r][c] == 1:
                    fresh += 1
                if grid[r][c] == 2:
                    q.append([r, c])

        directions = [[0, 1], [0, -1], [1, 0], [-1, 0]]

        while q and fresh > 0:
            for i in range(len(q)):
                r, c = q.popleft()
                for dr, dc in directions:
                    row, col = dr + r, dc + c
                    # check if in bounds
                    if (
                        row < 0
                        or col < 0
                        or row == len(grid)
                        or col == len(grid[0])
                        or grid[row][col] != 1
                    ):
                        continue
                    grid[row][col] = 2
                    q.append([row, col])
                    fresh -= 1
            time += 1
        return time if fresh == 0 else -1
```