994-rotting-oranges¶

Description¶

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(Python)¶

class Solution:
    def orangesRotting(self, grid: List[List[int]]) -> int:
        r = len(grid)
        c = len(grid[0])

        fresh = 0

        q = deque([])

        visited = set()
        days = 0

        for i in range(r):
            for j in range(c):
                if grid[i][j] == 2:
                    q.append((i, j, 0))
                elif grid[i][j] == 1:
                    fresh += 1
        if not fresh:
            return 0

        while q:
            n = len(q)
            for _ in range(n):
                x, y, time = q.popleft()
                days = max(days, time)
                for neigh_x, neigh_y in [
                    (x + 1, y),
                    (x - 1, y),
                    (x, y + 1),
                    (x, y - 1),
                ]:
                    if 0 <= neigh_x <= r - 1 and 0 <= neigh_y <= c - 1:
                        if (neigh_x, neigh_y) not in visited and grid[neigh_x][
                            neigh_y
                        ] == 1:
                            fresh -= 1
                            visited.add((neigh_x, neigh_y))
                            q.append((neigh_x, neigh_y, time + 1))

        return -1 if fresh else days