# O(MN) running time, O(1) space solution.

• Although this question doesn't really make sense. I even asked myself was it really from Google...lol...

'''

``````def gameOfLife(self, board):

def check(i,j):
m,n = len(board), len(board[0])
count = 0
if i-1 >= 0 and j-1 >= 0:
count += board[i-1][j-1][0]
if i-1 >= 0:
count += board[i-1][j][0]
if i-1 >= 0 and j+1 < n:
count += board[i-1][j+1][0]
if j-1 >= 0:
count += board[i][j-1][0]
if j+1 < n:
count += board[i][j+1][0]
if i+1 < m and j-1 >= 0:
count += board[i+1][j-1][0]
if i+1 < m:
count += board[i+1][j][0]
if i+1 < m and j+1 < n:
count += board[i+1][j+1][0]

if board[i][j][0] == 1:
if count < 2 or count > 3:
return 0
else:
return 1
else:
if count == 3:
return 1
else:
return 0

for i in xrange(len(board)):
for j in xrange(len(board[0])):
board[i][j] = [board[i][j]]

for i in xrange(len(board)):
for j in xrange(len(board[0])):
board[i][j].append(check(i,j))

for i in xrange(len(board)):
for j in xrange(len(board[0])):
board[i][j] = board[i][j][1]
``````

'''

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