--- title: '994. Rotting Oranges' description: 'You are given an m x n grid where each cell can have one of three values:' icon: dot topics: - { name: "Array", slug: "array" } - { name: "Breadth-First Search", slug: "breadth-first-search" } - { name: "Matrix", slug: "matrix" } sidebar: label: 'Rotting Oranges' badge: 'Medium' --- ### 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 ```