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77 lines
2.1 KiB
Plaintext
77 lines
2.1 KiB
Plaintext
---
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title: '994. Rotting Oranges'
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description: 'You are given an m x n grid where each cell can have one of three values:'
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icon: dot
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topics:
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- { name: "Array", slug: "array" }
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- { name: "Breadth-First Search", slug: "breadth-first-search" }
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- { name: "Matrix", slug: "matrix" }
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issue: "https://github.com/prdlk/leetcode/issues/137"
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sidebar:
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label: 'Rotting Oranges'
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badge: 'Medium'
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---
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### Example 1:
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- Input: `grid = [[2,1,1],[1,1,0],[0,1,1]]`
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- Output: `4`
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### Example 2:
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- Input: `grid = [[2,1,1],[0,1,1],[1,0,1]]`
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- Output: `-1`
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- Explanation: The orange in the bottom left corner (row `2`, column `0`) is never rotten, because rotting only happens 4-directionally.
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### Example 3:
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- Input: `grid = [[0,2]]`
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- Output: `0`
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- Explanation: Since there are already no fresh oranges at minute `0`, the answer is just `0`.
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### Constraints:
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- `m == grid.length`
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- `n == grid[i].length`
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- `1 <= m, n <= 10`
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- grid[i][j] is 0, 1, or 2.
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## Solution
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```py
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from collections import deque
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class Solution:
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def orangesRotting(self, grid: list[list[int]]) -> int:
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q = deque()
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time, fresh = 0, 0
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rows, cols = len(grid), len(grid[0])
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for r in range(rows):
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for c in range(cols):
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if grid[r][c] == 1:
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fresh += 1
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if grid[r][c] == 2:
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q.append([r, c])
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directions = [[0, 1], [0, -1], [1, 0], [-1, 0]]
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while q and fresh > 0:
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for i in range(len(q)):
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r, c = q.popleft()
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for dr, dc in directions:
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row, col = dr + r, dc + c
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# check if in bounds
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if (
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row < 0
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or col < 0
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or row == len(grid)
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or col == len(grid[0])
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or grid[row][col] != 1
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):
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continue
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grid[row][col] = 2
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q.append([row, col])
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fresh -= 1
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time += 1
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return time if fresh == 0 else -1
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```
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