Game of Life | Problems

The board is made up of an m x n grid of cells, where each cell is either alive (1) or dead (0). Each cell interacts with its eight neighbors (horizontal, vertical, diagonal) using the rules of Conway's Game of Life. Given the current state of the board, return the next state.

Sample Input

board = [[0,1,0],[0,0,1],[1,1,1],[0,0,0]]

Sample Output

[[0,0,0],[1,0,1],[0,1,1],[0,1,0]]

Constraints

1 <= m, n <= 25
board[i][j] is 0 or 1

Test Cases

Case 1

Args: [[[0,1,0],[0,0,1],[1,1,1],[0,0,0]]] Expected: [[0,0,0],[1,0,1],[0,1,1],[0,1,0]]

Topics

Array Two Pointers

To update in-place, encode transitions with extra states: 2 means alive → dead, 3 means dead → alive. Count live neighbors using original states (values 1 or 2), then do a final pass to map states back to 0/1.

function gameOfLife(board) {
  const m = board.length, n = board[0].length;
  const dirs = [-1, 0, 1];

  for (let i = 0; i < m; i++) {
    for (let j = 0; j < n; j++) {
      let live = 0;
      for (const di of dirs) {
        for (const dj of dirs) {
          if (di === 0 && dj === 0) continue;
          const ni = i + di, nj = j + dj;
          if (ni >= 0 && ni < m && nj >= 0 && nj < n) {
            if (board[ni][nj] === 1 || board[ni][nj] === 2) live++;
          }
        }
      }
      if (board[i][j] === 1 && (live < 2 || live > 3)) board[i][j] = 2;
      if (board[i][j] === 0 && live === 3) board[i][j] = 3;
    }
  }

  for (let i = 0; i < m; i++) {
    for (let j = 0; j < n; j++) {
      if (board[i][j] === 2) board[i][j] = 0;
      if (board[i][j] === 3) board[i][j] = 1;
    }
  }
}

Time: O(m × n) Space: O(1)

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