Rotting oranges

class Solution:
    def __init__(self):
        self.visited = set()
        self.directions = [(0, 1), (0, -1), (1, 0), (-1, 0)]
        self.start = []
        self.totalCount = 0
        self.time = 0

    def bfs(self, grid):
        q = collections.deque(self.start)

        while q:
            size = len(q)
            for i in range(size):
                x, y, self.time = q.popleft()
                for dx, dy in self.directions:
                    newx = x + dx
                    newy = y + dy
                    if 0 <= newx < len(grid) and 0 <= newy < len(grid[0]):
                        if (newx, newy) not in self.visited and grid[newx][newy] == 1:
                            q.append((newx, newy, self.time + 1))
                            self.visited.add((newx, newy))
                            grid[newx][newy] = 2


    def orangesRotting(self, grid: List[List[int]]) -> int:
        for i in range(len(grid)):
            for j in range(len(grid[0])):
                if grid[i][j] == 2:
                    self.start.append((i, j, 0))
                    self.visited.add((i, j))
                if grid[i][j] == 1:
                    self.totalCount += 1
        self.bfs(grid)

        for row in grid:
            if 1 in row:
                return -1

        return self.time

Rotting Oranges

Difficulty: Medium

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.