> For the complete documentation index, see [llms.txt](https://jaywin.gitbook.io/leetcode/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://jaywin.gitbook.io/leetcode/solutions/0289-game-of-life.md). # 0289. Game of Life ## Description According to [Wikipedia's article](https://en.wikipedia.org/wiki/Conway%27s_Game_of_Life): "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](https://en.wikipedia.org/wiki/Moore_neighborhood) (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 is created by applying the above rules simultaneously to every cell in the current state, where births and deaths occur simultaneously. Given the current state of the `m x n` grid `board`, return *the next state*. **Example 1:** ![](https://assets.leetcode.com/uploads/2020/12/26/grid1.jpg) ``` **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:** ![](https://assets.leetcode.com/uploads/2020/12/26/grid2.jpg) ``` **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? ## ac1: O(mn) space ```java class Solution { private final List DIRECTIONS = Arrays.asList( new int[] {1, 0}, new int[] {-1, 0}, new int[] {0, 1}, new int[] {0, -1}, new int[] {-1, -1}, new int[] {-1, 1}, new int[] {1, -1}, new int[] {1, 1} ); public void gameOfLife(int[][] board) { // edge case if (board.length == 0 || board[0].length == 0) return; int rows = board.length; int columns = board[0].length; int[][] copy = new int[rows][columns]; // first pass, mark in a new board for (int row = 0; row < rows; row++) { for (int col = 0; col < columns; col++) { int lives = countLives(board, row, col); if (lives < 2 || lives > 3) { copy[row][col] = 0; } else if (lives == 3) { copy[row][col] = 1; } else if (lives == 2) { if (board[row][col] == 1) copy[row][col] = 1; else copy[row][col] = 0; } } } // 2nd pass, modify base on new board for (int row = 0; row < rows; row++) { for (int col = 0; col < columns; col++) { board[row][col] = copy[row][col]; } } } private int countLives(int[][] board, int row, int col) { int lives = 0; for (int[] d : DIRECTIONS) { int r = row + d[0]; int c = col + d[1]; if (r >= board.length || r < 0 || c >= board[0].length || c < 0) continue; // out of boundary lives += board[r][c]; } return lives; } } ``` ## ac2: in-place ```java class Solution { private int[][] DIRECTIONS = new int[][]{{-1, -1}, {-1, 0}, {-1, 1}, {0, -1}, {0, 1}, {1, -1}, {1, 0}, {1, 1}}; public void gameOfLife(int[][] board) { // edge case if (board.length == 0 || board[0].length == 0) return; int rows = board.length; int columns = board[0].length; // int[][] copy = new int[rows][columns]; // first pass, mark in a new board for (int row = 0; row < rows; row++) { for (int col = 0; col < columns; col++) { int lives = countLives(board, row, col); if (lives < 2 || lives > 3) { // board[row][col] = 0; board[row][col] = board[row][col] == 1 ? 10 : 0; } else if (lives == 3) { // board[row][col] = 1; board[row][col] = board[row][col] == 1 ? 11 : 2; } else if (lives == 2) { board[row][col] = board[row][col] == 1 ? 11 : 0; } } } // 2nd pass, modify base on new board for (int row = 0; row < rows; row++) { for (int col = 0; col < columns; col++) { if (board[row][col] == 2 || board[row][col] == 11) { board[row][col] = 1; } else { board[row][col] = 0; } } } } private int countLives(int[][] board, int row, int col) { int lives = 0; for (int[] d : DIRECTIONS) { int r = row + d[0]; int c = col + d[1]; if (r >= board.length || r < 0 || c >= board[0].length || c < 0) continue; // out of boundary if (board[r][c] == 1 || board[r][c] == 11 || board[r][c] == 10) { lives++; } } return lives; } } ``` --- # Agent Instructions This documentation is published with GitBook. GitBook is the documentation platform designed so that both humans and AI agents can read, navigate, and reason over technical content effectively. Learn more at gitbook.com. ## Querying This Documentation If you need additional information that is not directly available in this page, you can query the documentation dynamically by asking a question. Perform an HTTP GET request on the current page URL with the `ask` query parameter: ``` GET https://jaywin.gitbook.io/leetcode/solutions/0289-game-of-life.md?ask= ``` The question should be specific, self-contained, and written in natural language. The response will contain a direct answer to the question and relevant excerpts and sources from the documentation. Use this mechanism when the answer is not explicitly present in the current page, you need clarification or additional context, or you want to retrieve related documentation sections.