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94 lines
2.3 KiB
Python
94 lines
2.3 KiB
Python
"""
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994. Rotting Oranges
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Difficulty: Medium
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https://leetcode.com/problems/rotting-oranges/
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──────────────────────────────────────────────────
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You are given an m x n grid where each cell can have one of three
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values:
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• 0 representing an empty cell,
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• 1 representing a fresh orange, or
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• 2 representing a rotten orange.
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Every minute, any fresh orange that is 4-directionally adjacent to a
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rotten orange becomes rotten.
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Return the minimum number of minutes that must elapse until no cell
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has a fresh orange. If this is impossible, return -1.
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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)
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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,
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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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"""
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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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