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994.RottingOranges.py
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994.RottingOranges.py
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"""
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:
Input: [[2,1,1],[1,1,0],[0,1,1]]
Output: 4
Example:
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-directiona
Note:
1. 1 <= grid.length <= 10
2. 1 <= grid[0].length <= 10
3. grid[i][j] is only 0, 1, or 2.
"""
#Difficulty: Medium
#303 / 303 test cases passed.
#Runtime: 52 ms
#Memory Usage: 13.7 MB
#Runtime: 52 ms, faster than 84.00% of Python3 online submissions for Rotting Oranges.
#Memory Usage: 13.7 MB, less than 91.03% of Python3 online submissions for Rotting Oranges.
class Solution:
def orangesRotting(self, grid: List[List[int]]) -> int:
minutes = 0
self.rows = len(grid)
self.cols = len(grid[0])
self.rotten_value = 2
self.rotting = True
while self.rotting and self.checkForFreshOranges(grid):
self.rotting = False
for i in range(self.rows):
for j in range(self.cols):
if grid[i][j] == self.rotten_value:
self.rottenOrange(grid, i - 1, j)
self.rottenOrange(grid, i, j - 1)
self.rottenOrange(grid, i, j + 1)
self.rottenOrange(grid, i + 1, j)
if self.rotting:
minutes += 1
self.rotten_value += 1
if not self.rotting:
return -1
return minutes
def rottenOrange(self, grid, i, j):
if i < 0 or i >= len(grid) or j < 0 or j >= len(grid[i]):
return
if grid[i][j] == 1:
grid[i][j] += self.rotten_value
self.rotting = True
def checkForFreshOranges(self, grid):
for i in range(self.rows):
for j in range(self.cols):
if grid[i][j] == 1:
return True
return False