# Program for Conway’s Game Of Life | Set 2

• Difficulty Level : Medium
• Last Updated : 07 Apr, 2022

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:

1. Any live cell with fewer than two live neighbors dies due to under-population.
2. Any live cell with two or three live neighbors lives on to the next generation.
3. Any live cell with more than three live neighbors dies due to overpopulation.
4. 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
0000000000
Output:
0000000000
0001100000
0001100000
0000000000
0000000000
Explanation:
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
00000
Output:
00000
00100
00100
00000
Explanation:
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.

1. Traverse the whole grid.
2. 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.
3. 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.

Below is the implementation of the above approach:

## C++

 // C++ implementation of// the above approach #include using namespace std; void print(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;    }}       // Function to check if// index are not out of gridstatic bool save(vector > grid,int row, int col){    return (grid.size() > row && grid[0].size() > col && row >= 0 && col >= 0);}    // Function to get New Generationvoid solve(vector >& grid){      int p = grid.size(),        q = grid[0].size();      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 codeint main (){  vector> 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);  return 0;}

## Java

 // Java program for// the above approachimport 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 codepublic 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

## Python3

 # Python3 implementation of# the above approach def print2dArr(grid):    p = len(grid)    q = len(grid[0])     for i in range(p):        for j in range(q):            print(grid[i][j], end='')         print()  # Function to check if# index are not out of griddef save(grid, row, col):    return (len(grid) > row and len(grid[0]) > col and row >= 0 and col >= 0) # Function to get New Generation  def solve(grid):    p = len(grid)    q = len(grid[0])    u = [1, -1, 0, 1, -1, 0, 1, -1]    v = [0, 0, -1, -1, -1, 1, 1, 1]    for i in range(p):        for j in range(q):            # IF the initial value            # of the grid(i, j) is 1            if (grid[i][j] > 0):                for k in range(8):                    if (save(grid, i + u[k], j + v[k]) and grid[i + u[k]][j + v[k]] > 0):                        # If initial value > 0,                        # just increment it by 1                        grid[i][j] += 1             # IF the initial value            # of the grid(i, j) is 0            else:                for k in range(8):                    if (save(grid, i + u[k], j + v[k]) and grid[i + u[k]][j + v[k]] > 0):                        # If initial value <= 0                        # just decrement it by 1                        grid[i][j] -= 1    # Generating new Generation.    # Now the magnitude of the    # grid will represent number    # of neighbours    for i in range(p):        for j in range(q):            # 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                elif (grid[i][j] <= 4):                    grid[i][j] = 1                 # Since Any live cell with                # > 3 live neighbors dies                elif (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 codeif __name__ == '__main__':    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    print2dArr(grid)

## Javascript



Output:
0000000000
0001100000
0001100000
0000000000
0000000000

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

My Personal Notes arrow_drop_up