Scale Engineer

Game of Life

Med
#0276Time: O(m*n), where m and n are the dimensions of the boardSpace: O(m*n) to store the next state grid8 companies
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Prompt

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):

  1. Any live cell with fewer than two live neighbors dies as if caused by 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, as if by over-population.
  4. Any dead cell with exactly three live neighbors becomes a live cell, as if by reproduction.

The next state of the board is determined by applying the above rules simultaneously to every cell in the current state of the m x n grid board. In this process, births and deaths occur simultaneously.

Given the current state of the board, update the board to reflect its next state.

Note that you do not need to return anything.

 

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?

Approaches

2 approaches with complexity analysis and trade-offs.

The most straightforward approach is to create a new grid to store the next state while processing the current grid.

Algorithm

  1. Create a new m x n grid nextState
  2. For each cell (i,j) in the board:
    • Count live neighbors in all 8 directions
    • Apply Game of Life rules to determine next state
    • Store result in nextState[i][j]
  3. Copy nextState back to original board

Walkthrough

In this approach, we create a new m x n grid to store the next state of the board. For each cell in the current board, we:

  1. Count the number of live neighbors using the 8 adjacent cells
  2. Apply the Game of Life rules to determine the next state
  3. Store the result in the new grid
  4. Finally, copy the new grid back to the original board

Here's the implementation:

public void gameOfLife(int[][] board) {    int m = board.length;    int n = board[0].length;    int[][] nextState = new int[m][n];        // Define the 8 directions for neighbors    int[] dx = {-1, -1, -1, 0, 0, 1, 1, 1};    int[] dy = {-1, 0, 1, -1, 1, -1, 0, 1};        // Process each cell    for (int i = 0; i < m; i++) {        for (int j = 0; j < n; j++) {            int liveNeighbors = 0;                        // Count live neighbors            for (int k = 0; k < 8; k++) {                int newX = i + dx[k];                int newY = j + dy[k];                                if (newX >= 0 && newX < m && newY >= 0 && newY < n) {                    liveNeighbors += board[newX][newY];                }            }                        // Apply rules            if (board[i][j] == 1) {                if (liveNeighbors < 2 || liveNeighbors > 3) {                    nextState[i][j] = 0;                } else {                    nextState[i][j] = 1;                }            } else {                if (liveNeighbors == 3) {                    nextState[i][j] = 1;                }            }        }    }        // Copy back to original board    for (int i = 0; i < m; i++) {        for (int j = 0; j < n; j++) {            board[i][j] = nextState[i][j];        }    }}

Complexity

Time

O(m*n), where m and n are the dimensions of the board

Space

O(m*n) to store the next state grid

Trade-offs

Pros

  • Simple to understand and implement

  • No risk of mixing old and new states

  • Easy to debug

Cons

  • Uses extra space proportional to the board size

  • Requires additional copy operation at the end

  • Not memory efficient for large boards

Solutions

public class Solution { public void GameOfLife ( int [][] board ) { int m = board . Length ; int n = board [ 0 ]. Length ; for ( int i = 0 ; i < m ; ++ i ) { for ( int j = 0 ; j < n ; ++ j ) { int live = - board [ i ][ j ]; for ( int x = i - 1 ; x <= i + 1 ; ++ x ) { for ( int y = j - 1 ; y <= j + 1 ; ++ y ) { if ( x >= 0 && x < m && y >= 0 && y < n && board [ x ][ y ] > 0 ) { ++ live ; } } } if ( board [ i ][ j ] == 1 && ( live < 2 || live > 3 )) { board [ i ][ j ] = 2 ; } if ( board [ i ][ j ] == 0 && live == 3 ) { board [ i ][ j ] = - 1 ; } } } for ( int i = 0 ; i < m ; ++ i ) { for ( int j = 0 ; j < n ; ++ j ) { if ( board [ i ][ j ] == 2 ) { board [ i ][ j ] = 0 ; } if ( board [ i ][ j ] == - 1 ) { board [ i ][ j ] = 1 ; } } } } }

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