# 289 Game of Life

According to the [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."

Given a board with m by n cells, each cell has an initial state live(1) or dead(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.

Write a function to compute the next state (after one update) of the board given its current state.

**Follow up**:

1. Could you solve it in-place? Remember that the board needs to be updated at the same time: You cannot update some cells first and then use their updated values to update other cells.
2. 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 the border of the array. How would you address these problems?

**Credits:**\
Special thanks to[@jianchao.li.fighter](https://leetcode.com/discuss/user/jianchao.li.fighter)for adding this problem and creating all test cases.

**The Idea:** One way to achieve constant space performance time is to do to passes through the matrix. In the first pass, we count the number of live cells, and mark the cell appropriately if the cell transitions to either dead or alive. I gave to distinctive markers: (2) indicating dead to alive, and (3) live to dead. In maintain accuracy of the model, we then also have to treat (2 or 0) as dead and (3 or 0) as alive. In our second pass through the matrix, we map all (3) cells to 0 and (2) cells to 1.

**Complexity:** O(2\*n\*m) time and O(1) space

```python
class Solution:
    def gameOfLife(self, board):
        """
        :type board: List[List[int]]
        : (0) - dead
        : (1) - live
        : (2) - dead to live
        : (3) - live to dead
        :rtype: void Do not return anything, modify board in-place instead.
        """

        def in_board(r, c, R, C):
            return r >= 0 and c >= 0 and c < C and r < R

        neighbors = [(1, 0), (0, 1), (-1, 0), (0, -1), (1, 1), (1, -1), (-1, 1), (-1, -1)]
        for i, row in enumerate(board):
            for j, cell in enumerate(row):
                count_live = 0
                for r, c in neighbors:
                    if in_board(i + r, j + c, len(board), len(board[0])) and (board[i + r][j + c] == 1 or board[i + r][j + c] == 3):
                        count_live += 1
                # Any live cell with fewer than two live neighbors dies, as if caused by under-population
                # Any live cell with more than three live neighbors dies, as if by over-population
                if (count_live < 2 or count_live > 3)and (cell == 1 or cell == 3):
                    board[i][j] = 3  # live to dead
                # Any dead cell with exactly three live neighbors becomes a live cell, as if by reproduction
                elif (count_live == 3) and (cell == 0 or cell == 2):
                    board[i][j] = 2  # dead to live

        for i, row in enumerate(board):
            for j, cell in enumerate(row):
                if cell == 3:
                    board[i][j] = 0
                elif cell == 2:
                    board[i][j] = 1
```


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