LeetCode
LeetCode 994 Rotting Oranges - Medium
994. Rotting Oranges -- Medium
994. Rotting Oranges — Medium
Problem
- Rotting Oranges -- Medium
In a given grid, each cell can have one of three values:
the value 0 representing an empty cell; the value 1 representing a fresh orange; the value 2 representing a rotten orange. Every minute, any fresh orange that is adjacent (4-directionally) 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 instead.
Example 1: Input: [[2,1,1],[1,1,0],[0,1,1]] Output: 4
Example 2: Input: [[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: [[0,2]] Output: 0 Explanation: Since there are already no fresh oranges at minute 0, the answer is just 0.
Note: 1 <= grid.length <= 10 1 <= grid[0].length <= 10 grid[i][j] is only 0, 1, or 2.
Solution
class Solution:
def orangesRotting(self, grid: List[List[int]]) -> int:
queue = collections.deque()
fresh_oranges = 0
ROWS, COLS = len(grid), len(grid[0])
for r in range(ROWS):
for c in range(COLS):
if grid[r][c] == 2:
queue.append((r, c))
elif grid[r][c] == 1:
fresh_oranges += 1
queue.append((-1, -1)) # dummy to mark the level
minutes_elapsed = -1
directions = [(-1, 0), (0, 1), (1, 0), (0, -1)]
while queue:
row, col = queue.popleft()
if row == -1: # finish the round
minutes_elapsed += 1
if queue: #to avoid the endless loop
queue.append((-1, -1))
else:
for d in directions:
neighbor_row, neighbor_col = row + d[0], col + d[1]
if ROWS > neighbor_row >= 0 and COLS > neighbor_col >= 0:
if grid[neighbor_row][neighbor_col] == 1:
grid[neighbor_row][neighbor_col] = 2
fresh_oranges -= 1
queue.append((neighbor_row, neighbor_col))
return minutes_elapsed if fresh_oranges == 0 else -1