Given a binary **grid[ ][ ]** of size **N*M**, with each cell containing either 0 or 1, where 1 represents an **alive cell** and the 0 represents a **dead cell**. The task is to generate the next generation of cells based on the following rules:

- Any live cell with fewer than two live neighbors dies due to 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 due to overpopulation.
- Any dead cell with exactly three live neighbors becomes a live cell by reproduction.

Here, the **neighbor** of a cell includes its adjacent cells as well as diagonal ones, so for each cell, a total of **8** neighbors are there.**Examples:**

Input:N = 5, M = 10

0000000000

0001100000

0000100000

0000000000

0000000000Output:

0000000000

0001100000

0001100000

0000000000

0000000000Explanation:

The cells at (1, 3), (1, 4) and (2, 4) are alive and have 2 neighbors,

So, according to Rule 2, they live on to the next generation.

The cell at (2, 3) is dead and has exactly 3 neighbors,

So, according to Rule 4, it becomes a live cell.Input:N = 4, M = 5

00000

01100

00010

00000Output:

00000

00100

00100

00000Explanation:

The cells at (1, 1) and (2, 3) have only 1 neighbor,

So, according to Rule 1, they die.

The cell at (1, 2) is alive and has 2 neighbors,

So, according to Rule 2, it lives on to the next generation.

The cell at (2, 2) is dead and has exactly 3 neighbors,

So, according to Rule 4, it becomes a live cell.

**Naive Approach:**

We have already discussed an approach to this problem in Program for Conway’s Game Of Life | Set 1. In this approach, an extra grid **future[ ][ ]** of size **N*M** is created to store the next generation of cells. **Time complexity:** O(N*M)**Auxiliary Space:** O(N*M)**Efficient Approach:**

A *space-optimized* approach is possible for this problem, which uses no extra space. For every **alive cell**, increment its value by 1 each time we encounter an alive neighbor for it. And, for every **dead cell**, decrement its value by 1 each time we encounter an alive neighbor for it. Now, if the value in a cell is smaller than 0, it means it is a dead cell and if the value is greater than 0, it is an alive cell, also the **magnitude** of value in a cell will represent the **number of its alive neighbors**. In this way, an extra grid will not be required and a space-optimized solution is obtained. Detailed steps of this approach are as follows:

For the present grid position, we will look at all the neighbors.

- Traverse the whole grid.
- If the value of the neighbor is
**>= 1**(i.e. initial value of the neighbor was 1)- If, the current cell is
**>= 1**, just**increment**the current cell value by 1.

Now, grid [ i ][ j ] > 0 will imply that initial grid [ i ][ j ] == 1. - Else, the current cell is
**<= 0**, just**decrement**the current cell value by 1.

Now, grid [ i ][ j ] <= 0 will imply that initial grid [ i ][ j ] == 0.

- If, the current cell is
- Now, generate the required new generation grid using the modified grid.
- If the current cell value is
**> 0**, there are 3 possible cases:- If the value is < 3 then change it to 0.
- Else if the value is <= 4 then change it to 1.
- Else change the value to 0.

- Else, the current cell value is
**<= 0**.- If the value is 3 then change it to 1.
- Else change it to 0.

- If the current cell value is

Below is the implementation of the above approach:

## C++

`// C++ implementation of` `// the above approach` `#include` `using` `namespace` `std;` `class` `GameOfLife {` `public` `:` ` ` `// Function to display the grid` ` ` `void` `print(vector<vector > grid);` ` ` `// Function to check if` ` ` `// index are not out of grid` ` ` `bool` `save(vector<vector > grid,` ` ` `int` `row, ` `int` `col);` ` ` `// Function to get New Generation` ` ` `void` `solve(vector<vector >& grid);` `};` `void` `GameOfLife::print(` ` ` `vector<vector > grid)` `{` ` ` `int` `p = grid.size(),` ` ` `q = grid[0].size();` ` ` `for` `(` `int` `i = 0; i < p; i++) {` ` ` `for` `(` `int` `j = 0; j < q; j++) {` ` ` `cout << grid[i][j];` ` ` `}` ` ` `cout << endl;` ` ` `}` `}` `bool` `GameOfLife::save(` ` ` `vector<vector > grid,` ` ` `int` `row, ` `int` `col)` `{` ` ` `return` `(grid.size() > row` ` ` `&& grid[0].size() > col` ` ` `&& row >= 0` ` ` `&& col >= 0);` `}` `// Possible neighboring indexes` `int` `u[] = { 1, -1, 0, 1, -1, 0, 1, -1 };` `int` `v[] = { 0, 0, -1, -1, -1, 1, 1, 1 };` `void` `GameOfLife::solve(` ` ` `vector<vector >& grid)` `{` ` ` `int` `p = grid.size(),` ` ` `q = grid[0].size();` ` ` `for` `(` `int` `i = 0; i < p; i++) {` ` ` `for` `(` `int` `j = 0; j` |

## Java

`// Java program for` `// the above approach` `import` `java.util.*;` `import` `java.lang.*;` `class` `GFG{` ` ` `static` `void` `print(` `int` `[][] grid)` `{` ` ` `int` `p = grid.length,` ` ` `q = grid[` `0` `].length;` ` ` `for` `(` `int` `i = ` `0` `; i < p; i++)` ` ` `{` ` ` `for` `(` `int` `j = ` `0` `; j < q; j++)` ` ` `{` ` ` `System.out.print(grid[i][j]);` ` ` `}` ` ` `System.out.println();;` ` ` `}` `}` `static` `boolean` `save(` `int` `[][] grid,` ` ` `int` `row, ` `int` `col)` `{` ` ` `return` `(grid.length > row &&` ` ` `grid[` `0` `].length > col &&` ` ` `row >= ` `0` `&& col >= ` `0` `);` `}` `static` `void` `solve(` `int` `[][] grid)` `{` ` ` `int` `p = grid.length,` ` ` `q = grid[` `0` `].length;` ` ` `// Possible neighboring` ` ` `// indexes` ` ` `int` `u[] = {` `1` `, -` `1` `, ` `0` `, ` `1` `,` ` ` `-` `1` `, ` `0` `, ` `1` `, -` `1` `};` ` ` `int` `v[] = {` `0` `, ` `0` `, -` `1` `, -` `1` `,` ` ` `-` `1` `, ` `1` `, ` `1` `, ` `1` `};` ` ` ` ` `for` `(` `int` `i = ` `0` `; i < p; i++)` ` ` `{` ` ` `for` `(` `int` `j = ` `0` `; j < q; j++)` ` ` `{` ` ` `// IF the initial value` ` ` `// of the grid(i, j) is 1` ` ` `if` `(grid[i][j] > ` `0` `)` ` ` `{` ` ` `for` `(` `int` `k = ` `0` `; k < ` `8` `; k++)` ` ` `{` ` ` `if` `(save(grid, i + u[k],` ` ` `j + v[k]) &&` ` ` `grid[i + u[k]][j + v[k]] > ` `0` `)` ` ` `{` ` ` `// If initial value > 0,` ` ` `// just increment it by 1` ` ` `grid[i][j]++;` ` ` `}` ` ` `}` ` ` `}` ` ` `// IF the initial value` ` ` `// of the grid(i, j) is 0` ` ` `else` ` ` `{` ` ` `for` `(` `int` `k = ` `0` `; k < ` `8` `; k++)` ` ` `{` ` ` `if` `(save(grid, i + u[k],` ` ` `j + v[k]) &&` ` ` `grid[i + u[k]][j + v[k]] > ` `0` `)` ` ` `{` ` ` `// If initial value <= 0` ` ` `// just decrement it by 1` ` ` `grid[i][j]--;` ` ` `}` ` ` `}` ` ` `}` ` ` `}` ` ` `}` ` ` `// Generating new Generation.` ` ` `// Now the magnitude of the` ` ` `// grid will represent number` ` ` `// of neighbours` ` ` `for` `(` `int` `i = ` `0` `; i < p; i++)` ` ` `{` ` ` `for` `(` `int` `j = ` `0` `; j < q; j++)` ` ` `{` ` ` `// If initial value was 1.` ` ` `if` `(grid[i][j] > ` `0` `)` ` ` `{` ` ` `// Since Any live cell with` ` ` `// < 2 live neighbors dies` ` ` `if` `(grid[i][j] < ` `3` `)` ` ` `grid[i][j] = ` `0` `;` ` ` `// Since Any live cell with` ` ` `// 2 or 3 live neighbors live` ` ` `else` `if` `(grid[i][j] <= ` `4` `)` ` ` `grid[i][j] = ` `1` `;` ` ` `// Since Any live cell with` ` ` `// > 3 live neighbors dies` ` ` `else` `if` `(grid[i][j] > ` `4` `)` ` ` `grid[i][j] = ` `0` `;` ` ` `}` ` ` `else` ` ` `{` ` ` `// Since Any dead cell with` ` ` `// exactly 3 live neighbors` ` ` `// becomes a live cell` ` ` `if` `(grid[i][j] == -` `3` `)` ` ` `grid[i][j] = ` `1` `;` ` ` `else` ` ` `grid[i][j] = ` `0` `;` ` ` `}` ` ` `}` ` ` `}` `}` `// Driver code` `public` `static` `void` `main (String[] args)` `{` ` ` `int` `[][] grid = {{` `0` `, ` `0` `, ` `0` `, ` `0` `, ` `0` `,` ` ` `0` `, ` `0` `, ` `0` `, ` `0` `, ` `0` `},` ` ` `{` `0` `, ` `0` `, ` `0` `, ` `1` `, ` `1` `,` ` ` `0` `,` `0` `, ` `0` `, ` `0` `, ` `0` `},` ` ` `{` `0` `, ` `0` `, ` `0` `, ` `0` `, ` `1` `,` ` ` `0` `, ` `0` `, ` `0` `, ` `0` `, ` `0` `},` ` ` `{` `0` `, ` `0` `, ` `0` `, ` `0` `, ` `0` `,` ` ` `0` `, ` `0` `, ` `0` `, ` `0` `, ` `0` `},` ` ` `{` `0` `, ` `0` `, ` `0` `, ` `0` `, ` `0` `,` ` ` `0` `, ` `0` `, ` `0` `, ` `0` `, ` `0` `}};` ` ` ` ` `// Function to generate` ` ` `// New Generation inplace` ` ` `solve(grid);` ` ` `// Displaying the grid` ` ` `print(grid);` `}` `}` `// This code is contributed by offbeat` |

**Output:**

0000000000 0001100000 0001100000 0000000000 0000000000

**Time complexity:** O(N*M)**Auxiliary Space:** O(1)

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