# Python solution using a stack

• After coming up with this, I see people taken advantage of the expected values and using other parts to store the next state. Then going on to use a bitwise operations to get the the next state. That did not occur to me. I used a stack to keep track of the cell that needed to be updated and then updating after all cells are evaluated. This is not a solution for the infinite follow up question. Please let me know what you think.

``````class Solution(object):
def gameOfLife(self, board):
def getNeighbors(i,j, n, m, b):
neighbors = 0
# above
if i > 0:
if j > 0: neighbors += b[i-1][j-1]
if j+1 < m: neighbors += b[i-1][j+1]
neighbors += b[i-1][j]
# inline
if j > 0: neighbors += b[i][j-1]
if j+1 < m: neighbors += b[i][j+1]
# below
if i+1 < n:
neighbors += b[i+1][j]
if j > 0: neighbors += b[i+1][j-1]
if j+1 < m: neighbors += b[i+1][j+1]
return neighbors

n = len(board)
m = len(board[0])
stack = []
for i, row in enumerate(board):
for j, cell in enumerate(row):
neighbors = getNeighbors(i, j, n, m, board)
if cell:
if neighbors < 2 or neighbors > 3:
stack.append((i,j,0))
else:
if neighbors == 3:
stack.append((i,j,1))
while stack:
top = stack.pop()
board[top[0]][top[1]] = top[2]
``````

Thank you!

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