994. Rotting Oranges

Problem Link

Description

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You are given an m x n grid where each cell can have one of three values:

• 0 representing an empty cell,
• 1 representing a fresh orange, or
• 2 representing a rotten orange.

Every minute, any fresh orange that is 4-directionally adjacent 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.

 

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

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Python3

class Solution:
    def orangesRotting(self, grid: List[List[int]]) -> int:
        rows, cols = len(grid), len(grid[0])
        fresh = mins = 0
        queue = collections.deque()
        
        for x in range(rows):
            for y in range(cols):
                if grid[x][y] == 2:
                    queue.append((x, y))
                    
                elif grid[x][y] == 1:
                    fresh += 1
                    
        if fresh == 0: return 0
        
        while queue:
            n = len(queue)
            
            for _ in range(n):
                x, y = queue.popleft()
                
                for dx, dy in [(x + 1, y), (x - 1, y), (x, y + 1), (x, y - 1)]:
                    if 0 <= dx < rows and 0 <= dy < cols and grid[dx][dy] == 1:
                        grid[dx][dy] = 2
                        queue.append((dx, dy))
                        fresh -= 1
                        
            mins += 1
        
        return mins - 1 if fresh == 0 else -1