Problem description:

In a given grid, each cell can have one of three values:

the value 0 representing an empty cell;
the value 1 representing a fresh orange;
the value 2 representing a rotten orange.
Every minute, any fresh orange that is adjacent (4-directionally) 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 instead.

Example 1:

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Input: [[2,1,1],[1,1,0],[0,1,1]]
Output: 4

Example 2:

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Input: [[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:

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Input: [[0,2]]
Output: 0
Explanation:  Since there are already no fresh oranges at minute 0, the answer is just 0.

Note:

  1. 1 <= grid.length <= 10
  2. 1 <= grid[0].length <= 10
  3. grid[i][j] is only 0, 1, or 2.

Solution:

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class Solution:
    def orangesRotting(self, grid: List[List[int]]) -> int:
        # 1. find all rotten orange
        # 2. for each rotten orange, do bfs
          # 2.1 the rotten orange is added based on level
          # ex: at first, there're only 3 rotten orange, then just do 3 times BFS
          # 2.2 add new rotten orange to queue
        m = len(grid)
        if m == 0:
            return -1
        n = len(grid[0])
        
        queue = deque()
        fresh_orange = 0
        for i in range(m):
            for j in range(n):
                if grid[i][j] == 2:
                    queue.append((i, j))
                if grid[i][j] == 1:
                    fresh_orange += 1
        
        minute_passed = 0
        while queue and fresh_orange > 0:
            minute_passed += 1
            for _ in range(len(queue)):
                x, y = queue.popleft()
                for dx, dy in ((1, 0), (-1, 0), (0, 1), (0, -1)):
                    xx, yy = x + dx, y + dy
                    if xx < 0 or yy < 0 or xx == m or yy == n:
                        continue
                    if grid[xx][yy] == 0 or grid[xx][yy] == 2:
                        continue
                
                    grid[xx][yy] = 2
                    fresh_orange -= 1
                    queue.append((xx, yy))
        return minute_passed if fresh_orange == 0 else -1

time complexity: $O(mn)$, each cell is visited at least once

space complexity: $O(mn)$, in the worst case if all the oranges are rotten they will be added to the queue

reference:
https://leetcode.com/problems/rotting-oranges/discuss/563686/Python-Clean-and-Well-Explained-(faster-than-greater-90)

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