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.
Input: grid = [[2,1,1],[1,1,0],[0,1,1]] Output: 4
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.
Input: grid = [[0,2]] Output: 0 Explanation: Since there are already no fresh oranges at minute 0, the answer is just 0.
- m == grid.length
- n == grid[i].length
- 1 <= m, n <= 10
- grid[i][j] is 0, 1, or 2.
I solved this problem like this.
- We can solve this problem to bfs. Using deque!
class Solution: def orangesRotting(self, grid: List[List[int]]) -> int: dr, dc = [-1, 0, 1, 0], [0, 1, 0, -1] depth = 0 deq = deque() for r in range(len(grid)): for c in range(len(grid[r])): if grid[r][c] == 2: deq.append((r,c)) while deq: now_nodes = deq.copy() deq.clear() for node in now_nodes: r, c = node for idx in range(4): n_r, n_c = r + dr[idx], c + dc[idx] if not (0 <= n_r <= len(grid)-1) or not (0 <= n_c <= len(grid)-1): continue if grid[n_r][n_c] == 2: continue if grid[n_r][n_c] == 1: deq.append((n_r, n_c)) grid[n_r][n_c] = 2 continue if not deq: break depth += 1 for r in range(len(grid)): for c in range(len(grid[r])): if grid[r][c] == 1: return -1 return depth