To solve it in place, we use 2 bits to store 2 states:

```
[2nd bit, 1st bit] = [next state, current state]
- 00 dead (next) <- dead (current)
- 01 dead (next) <- live (current)
- 10 live (next) <- dead (current)
- 11 live (next) <- live (current)
```

- In the beginning, every cell is either
`00`

or`01`

. - Notice that
`1st`

state is independent of`2nd`

state. - Imagine all cells are instantly changing from the
`1st`

to the`2nd`

state, at the same time. - Let's count # of neighbors from
`1st`

state and set`2nd`

state bit. - Since every
`2nd`

state is by default dead, no need to consider transition`01 -> 00`

. - In the end, delete every cell's
`1st`

state by doing`>> 1`

.

For each cell's `1st`

bit, check the 8 pixels around itself, and set the cell's `2nd`

bit.

- Transition
`01 -> 11`

: when`board == 1`

and`lives >= 2 && lives <= 3`

. - Transition
`00 -> 10`

: when`board == 0`

and`lives == 3`

.

To get the current state, simply do

```
board[i][j] & 1
```

To get the next state, simply do

```
board[i][j] >> 1
```

Hope this helps!

```
public void gameOfLife(int[][] board) {
if (board == null || board.length == 0) return;
int m = board.length, n = board[0].length;
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
int lives = liveNeighbors(board, m, n, i, j);
// In the beginning, every 2nd bit is 0;
// So we only need to care about when will the 2nd bit become 1.
if (board[i][j] == 1 && lives >= 2 && lives <= 3) {
board[i][j] = 3; // Make the 2nd bit 1: 01 ---> 11
}
if (board[i][j] == 0 && lives == 3) {
board[i][j] = 2; // Make the 2nd bit 1: 00 ---> 10
}
}
}
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
board[i][j] >>= 1; // Get the 2nd state.
}
}
}
public int liveNeighbors(int[][] board, int m, int n, int i, int j) {
int lives = 0;
for (int x = Math.max(i - 1, 0); x <= Math.min(i + 1, m - 1); x++) {
for (int y = Math.max(j - 1, 0); y <= Math.min(j + 1, n - 1); y++) {
lives += board[x][y] & 1;
}
}
lives -= board[i][j] & 1;
return lives;
}
```