Last updated on October 5th, 2024 at 05:30 pm

Here, We see ** Game of Life LeetCode Solution**. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.

*List of all LeetCode Solution*

*List of all LeetCode Solution*

## Topics

Array

## Companies

Dropbox, Google, Snapchat, TwoSigma

## Level of Question

Medium

**Game of Life LeetCode Solution**

## Table of Contents

**Problem Statement**

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]]

**1. Game of Life LeetCode Solution C++**

class Solution { public: 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; } };

**2. Game of Life LeetCode Solution Java**

class Solution { 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); if (board[i][j] == 1 && lives >= 2 && lives <= 3) { board[i][j] = 3; } if (board[i][j] == 0 && lives == 3) { board[i][j] = 2; } } } for (int i = 0; i < m; i++) { for (int j = 0; j < n; j++) { board[i][j] >>= 1; } } } 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; } }

**3. Game of Life LeetCode Solution JavaScript**

var gameOfLife = function(board) { if(board.length === 0){ return board; } var checkNeighbors = function(row, col){ var score = -board[row][col]; var r, c; for(r = row - 1; r <= row + 1; r++){ for(c = col - 1; c <= col + 1; c++){ if(typeof board[r] !== "undefined" && typeof board[r][c] !== "undefined"){ score += Math.abs(Math.floor(board[r][c])); } } } return score; }; var r, c; for(r = 0; r < board.length; r++){ for(c = 0; c < board[0].length; c++){ var score = checkNeighbors(r, c); if(board[r][c] === 1){ if(score < 2 || score > 3){ board[r][c] = -0.5; } } else if(board[r][c] === 0){ if(score === 3){ board[r][c] = 0.5; } } } } for(r = 0; r < board.length; r++){ for(c = 0; c < board[0].length; c++){ board[r][c] = Math.ceil(board[r][c]); } } };

**4. Game of Life LeetCode Solution Python**

class Solution(object): def gameOfLife(self, board): m, n = len(board), len(board[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(board[nr][nc]) == 1: livecount += 1 if board[i][j] == 1: if livecount < 2 or livecount > 3: board[i][j] = -1 else: if livecount == 3: board[i][j] = 2 for i in range(m): for j in range(n): if board[i][j] == 2: board[i][j] = 1 elif board[i][j] == -1: board[i][j] = 0