**Physical Address**

304 North Cardinal St.

Dorchester Center, MA 02124

According to Wikipedia’s article: “The **Game of Life**, also known simply as **Life**, is a cellular automaton devised by the British mathematician John Horton Conway in 1970.”

The board is made up of an `m x n`

grid of cells, where each cell has an initial state: **live** (represented by a `1`

) or **dead** (represented by a `0`

). Each cell interacts with its eight neighbors (horizontal, vertical, diagonal) using the following four rules (taken from the above Wikipedia article):

- Any live cell with fewer than two live neighbors dies as if caused by under-population.
- Any live cell with two or three live neighbors lives on to the next generation.
- Any live cell with more than three live neighbors dies, as if by over-population.
- Any dead cell with exactly three live neighbors becomes a live cell, as if by reproduction.

The next state is created by applying the above rules simultaneously to every cell in the current state, where births and deaths occur simultaneously. Given the current state of the `m x n`

grid `board`

, return *the next state*.

**Example 1:**

```
Input: board = [[0,1,0],[0,0,1],[1,1,1],[0,0,0]]
Output: [[0,0,0],[1,0,1],[0,1,1],[0,1,0]]
```

**Example 2:**

```
Input: board = [[1,1],[1,0]]
Output: [[1,1],[1,1]]
```

**Constraints:**

`m == board.length`

`n == board[i].length`

`1 <= m, n <= 25`

`board[i][j]`

is`0`

or`1`

.

**Follow up:**

- Could you solve it in-place? Remember that the board needs to be updated simultaneously: You cannot update some cells first and then use their updated values to update other cells.
- In this question, we represent the board using a 2D array. In principle, the board is infinite, which would cause problems when the active area encroaches upon the border of the array (i.e., live cells reach the border). How would you address these problems?

```
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;
}
```

```
void gameOfLife(vector<vector<int>>& board) {
int m = board.size(), n = m ? board[0].size() : 0;
for (int i=0; i<m; ++i) {
for (int j=0; j<n; ++j) {
int count = 0;
for (int I=max(i-1, 0); I<min(i+2, m); ++I)
for (int J=max(j-1, 0); J<min(j+2, n); ++J)
count += board[I][J] & 1;
if (count == 3 || count - board[i][j] == 3)
board[i][j] |= 2;
}
}
for (int i=0; i<m; ++i)
for (int j=0; j<n; ++j)
board[i][j] >>= 1;
}
```

```
class Solution:
def gameOfLife(self, b: List[List[int]]) -> None:
"""
Do not return anything, modify board in-place instead.
apply by using:
under-population: < 2
live to next generation: 2 or 3
over-population: > 3
reproduction: == 3
simultaneously: shoud not use DFS/BFS
\|/
- -
/|\
"""
# all new 0's denotes as -1, (1 ==> 0)
# all new 1's denotes as 2 (0 ==> 1)
m, n = len(b), len(b[0])
dirs = [[-1,-1],[-1,0],[-1,1],[0,1],[1,1],[1,0],[1,-1],[0,-1]]
for i in range(m):
for j in range(n):
livecount = 0
for r, c in dirs:
nr, nc = i + r, j + c
if 0 <= nr < m and 0 <= nc < n and abs(b[nr][nc]) == 1: # originally 1's
livecount += 1
if b[i][j] == 1:
if livecount < 2 or livecount > 3:
b[i][j] = -1
else:
if livecount == 3:
b[i][j] = 2
for i in range(m):
for j in range(n):
if b[i][j] == 2: b[i][j] = 1
elif b[i][j] == -1: b[i][j] = 0
```

In our experience, we suggest you solve this Game of Life LeetCode Solution and gain some new skills from Professionals completely free and we assure you will be worth it.

If you are stuck anywhere between any coding problem, just visit Queslers to get the Game of Life LeetCode Solution

I hope this Game of Life LeetCode Solution would be useful for you to learn something new from this problem. If it helped you then don’t forget to bookmark our site for more Coding Solutions.

This Problem is intended for audiences of all experiences who are interested in learning about Data Science in a business context; there are no prerequisites.

Keep Learning!

**More Coding Solutions >>**