LeetCode 289: Game of Life Solution

Master LeetCode problem 289 (Game of Life), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

289. Game of Life

Medium

Array Matrix Simulation

Problem Explanation

Explanation

The problem requires updating the given board based on the rules of the Game of Life. We need to update the board in-place, considering the next state of each cell simultaneously. To do this, we can follow the following steps:

Iterate through each cell in the board and calculate the number of live neighbors for each cell.
Based on the current state of the cell and the number of live neighbors, update the cell to its next state according to the rules.
To avoid updating cells and then using the updated values to update other cells, we can encode the next state in a different way. For example, we can use additional bits to represent the next state.

Time complexity: O(m*n) where m is the number of rows and n is the number of columns in the board. Space complexity: O(1) as we are updating the board in-place without using any additional data structures.

Solution Code

class Solution {
    public void gameOfLife(int[][] board) {
        int m = board.length;
        int n = board[0].length;

        int[][] directions = {{-1, -1}, {-1, 0}, {-1, 1}, {0, -1}, {0, 1}, {1, -1}, {1, 0}, {1, 1}};

        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                int liveNeighbors = 0;
                for (int[] dir : directions) {
                    int x = i + dir[0];
                    int y = j + dir[1];
                    if (x >= 0 && x < m && y >= 0 && y < n && (board[x][y] == 1 || board[x][y] == 2)) {
                        liveNeighbors++;
                    }
                }
                if (board[i][j] == 1 && (liveNeighbors < 2 || liveNeighbors > 3)) {
                    board[i][j] = 2; // cell dies
                } else if (board[i][j] == 0 && liveNeighbors == 3) {
                    board[i][j] = 3; // cell becomes live
                }
            }
        }

        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                board[i][j] = board[i][j] % 2;
            }
        }
    }
}

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