994. Rotting Oranges

Medium

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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.

class Solution:
    def orangesRotting(self, grid: List[List[int]]) -> int:
        row = len(grid)
        col = len(grid[0])
        dq = deque()
        time = 0
        fresh = set() # To Check if fresh is still there and fresh not updated twice
        for i in range(row):
            for j in range(col):
                if grid[i][j] == 1:
                    fresh.add((i,j))
                elif grid[i][j] == 2:
                    dq.append((i,j,0)) # Third thing is time Rotting orange has 0 time
        while dq:
            i,j,thistime = dq.popleft()
            time = max(time,thistime)
            for x,y in ((i-1,j), (i,j-1), (i+1,j), (i,j+1)):
                # not checking any boundary because we have already fresh indexes that we will check below
                if (x,y) in fresh:
                    dq.append((x,y,thistime+1))
                    fresh.remove((x,y))
        return -1 if fresh else time