LC289. Game of Life

Posted on | In |

Description

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
According to the 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."

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 (horizontal, vertical, diagonal) using the following four rules (taken from the above Wikipedia article):

Any live cell with fewer than two live neighbors dies, as if caused by under-population.
Any live cell with two or three live neighbors lives on to the next generation.
Any live cell with more than three live neighbors dies, as if by over-population..
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. The next state is created by applying the above rules simultaneously to every cell in the current state, where births and deaths occur simultaneously.

Example:

Input:
[
  [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]
]
Follow up:

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.
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?

This problem is kind of easy if we deal with it by O(n n) time complexity and O(n n) space complexity, however, according to the follow up, we should come up with a O(1) space solution.

Obviously, we have to modify the original matrix directly to get O(1) space, but if we change the cell before, we can not get correct number of live neighbors after that.

Therefore, we can use bit manipulation to handle it. The code is very easy to understand - we can use 2 bits to store the information, the 1st bit is whether it is live after 1st round, the 2nd bit is whether it is live before 1st round.

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
class Solution:
    def gameOfLife(self, board: List[List[int]]) -> None:
        if not board or not board[0]: return
        m, n = len(board), len(board[0])
        for i in range(m):
            for j in range(n):
                cnt = 0
                for di, dj in ((1, 0), (-1, 0), (0, 1), (0, -1), (1, 1), (-1, 1), (1, -1), (-1, -1)):
                    ii, jj = i + di, j + dj
                    if 0 <= ii < m and 0 <= jj < n:
                        cnt += board[ii][jj] & 1
                if board[i][j] == 1 and (cnt == 2 or cnt == 3):
                    board[i][j] = 3
                elif board[i][j] == 0 and cnt == 3:
                    board[i][j] = 2
        for i in range(m):
            for j in range(n):
                board[i][j] >>= 1
Donate comment here