# [leetcode] 994. Rotting Oranges _ Algorithm Problem Solve for python

## 1. Problem

994. Rotting Oranges

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.

## 2. Solution

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[0])-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