• Time:O(mn)
• Space:O(1)

## C++

``````class Solution {
public:
void gameOfLife(vector<vector<int>>& board) {
const int m = board.size();
const int n = board[0].size();

for (int i = 0; i < m; ++i)
for (int j = 0; j < n; ++j) {
int ones = 0;
for (int x = max(0, i - 1); x < min(m, i + 2); ++x)
for (int y = max(0, j - 1); y < min(n, j + 2); ++y)
ones += board[x][y] & 1;
// any live cell with 2 or 3 live neighbors
// lives on to the next generation
if (board[i][j] == 1 && (ones == 3 || ones == 4))
board[i][j] |= 0b10;
// any dead cell with exactly 3 live neighbors
// becomes a live cell, as if by reproduction
if (board[i][j] == 0 && ones == 3)
board[i][j] |= 0b10;
}

for (int i = 0; i < m; ++i)
for (int j = 0; j < n; ++j)
board[i][j] >>= 1;
}
};
``````

## JAVA

``````class Solution {
public void gameOfLife(int[][] board) {
final int m = board.length;
final int n = board[0].length;

for (int i = 0; i < m; ++i)
for (int j = 0; j < n; ++j) {
int ones = 0;
for (int x = Math.max(0, i - 1); x < Math.min(m, i + 2); ++x)
for (int y = Math.max(0, j - 1); y < Math.min(n, j + 2); ++y)
ones += board[x][y] & 1;
// any live cell with 2 or 3 live neighbors
// lives on to the next generation
if (board[i][j] == 1 && (ones == 3 || ones == 4))
board[i][j] |= 0b10;
// any dead cell with exactly 3 live neighbors
// becomes a live cell, as if by reproduction
if (board[i][j] == 0 && ones == 3)
board[i][j] |= 0b10;
}

for (int i = 0; i < m; ++i)
for (int j = 0; j < n; ++j)
board[i][j] >>= 1;
}
}
``````

## Python

``````class Solution:
def gameOfLife(self, board: List[List[int]]) -> None:
m = len(board)
n = len(board[0])

for i in range(m):
for j in range(n):
ones = 0
for x in range(max(0, i - 1), min(m, i + 2)):
for y in range(max(0, j - 1), min(n, j + 2)):
ones += board[x][y] & 1
# any live cell with 2 or 3 live neighbors
# lives on to the next generation
if board[i][j] == 1 and (ones == 3 or ones == 4):
board[i][j] |= 0b10
# any dead cell with exactly 3 live neighbors
# becomes a live cell, as if by reproduction
if board[i][j] == 0 and ones == 3:
board[i][j] |= 0b10

for i in range(m):
for j in range(n):
board[i][j] >>= 1
``````