# Easy state encoding to remember the old state as well as the new state, do not need bit manipulation

• define state change machine to remember the old state as well as the new result,
case 1 -> 2 from live to dead
case 2 -> 3 from live to live
case 3 -> 4 from live to dead
case 4 -> 5 from dead to live
ask for in place, then count costs time...
if ask for time efficiency, define a new 2D array for time saving

``````public void gameOfLife(int[][] board) {
if(board.length <= 0 && board[0].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 count = countNeighbor(board, i, j, m, n);
if(board[i][j] == 1){
if(count < 2){
board[i][j] = 2;
}else if(count == 2 || count == 3){
board[i][j] = 3;
}else{
board[i][j] = 4;
}
}else{
if(count == 3) board[i][j] = 5;
}
}
}

for(int i = 0; i < m; i++){
for(int j = 0; j < n; j++){
if(board[i][j] == 2){
board[i][j] = 0;
}else if(board[i][j] == 3){
board[i][j] = 1;
}else if(board[i][j] == 4){
board[i][j] = 0;
}else if(board[i][j] == 5){
board[i][j] = 1;
}
}
}
return;
}

public int countNeighbor(int[][] board, int i, int j, int m, int n){
int count = 0;
// board[i][j] in [1, 4] means it is live in old state
// board[i][j] in {0, 5} means it's dead in previous state
if(i > 0 && j > 0 && board[i-1][j-1] >= 1 && board[i-1][j-1] <= 4) count++;
if(i > 0 && board[i-1][j] >= 1 && board[i-1][j] <= 4) count++;
if(i > 0 && j < n-1 && board[i-1][j+1] >= 1 && board[i-1][j+1] <= 4) count++;
if(j > 0 && board[i][j-1] >= 1 && board[i][j-1] <= 4) count++;
if(j < n-1 && board[i][j+1] >= 1 && board[i][j+1] <= 4) count++;
if(i < m-1 && j < n-1 && board[i+1][j+1] >= 1 && board[i+1][j+1] <= 4) count++;
if(i < m-1 && board[i+1][j] >= 1 && board[i+1][j] <= 4) count++;
if(i < m-1 && j > 0 && board[i+1][j-1] >= 1 && board[i+1][j-1] <=4) count++;
return count;
}``````

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