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72_Rotten_Oranges
53 lines (52 loc) · 1.75 KB
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72_Rotten_Oranges
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class Solution:
#Function to find minimum time required to rot all oranges.
def orangesRotting(self, grid):
#Code here
n=len(grid)
m=len(grid[0])
rottend=[]
for i in range(n):
l=[]
for j in range(m):
l.append(2**32-1)
rottend.append(l)
x=[]
for i in range(n):
for j in range(m):
if grid[i][j]==2:
x.append([i,j])
rottend[i][j]=0
time=1
ans=0
while x:
new=[]
for [i,j] in x:
if i-1>=0 and grid[i-1][j]==1:
new.append([i-1,j])
rottend[i-1][j]=min(rottend[i-1][j],time)
# ans=max(ans,rottend[i-1][j])
grid[i-1][j]=2
if j-1>=0 and grid[i][j-1]==1:
new.append([i,j-1])
rottend[i][j-1]=min(rottend[i][j-1],time)
# ans=max(ans,rottend[i][j-1])
grid[i][j-1]=2
if i+1<n and grid[i+1][j]==1:
new.append([i+1,j])
rottend[i+1][j]=min(rottend[i+1][j],time)
# ans=max(ans,rottend[i+1][j])
grid[i+1][j]=2
if j+1<m and grid[i][j+1]==1:
new.append([i,j+1])
rottend[i][j+1]=min(rottend[i][j+1],time)
# ans=max(ans,rottend[i][j+1])
grid[i][j+1]=2
time+=1
x=new
for i in range(n):
for j in range(m):
if grid[i][j]==1:
return -1
if grid[i][j]==2:
ans=max(ans,rottend[i][j])
return ans