957. Prison Cells After N Days

https://leetcode.com/problems/prison-cells-after-n-days/

There are 8 prison cells in a row, and each cell is either occupied or vacant.

Each day, whether the cell is occupied or vacant changes according to the following rules:

• If a cell has two adjacent neighbors that are both occupied or both vacant, then the cell becomes occupied.

• Otherwise, it becomes vacant.

(Note that because the prison is a row, the first and the last cells in the row can't have two adjacent neighbors.)

We describe the current state of the prison in the following way: cells[i] == 1 if the i-th cell is occupied, else cells[i] == 0.

Given the initial state of the prison, return the state of the prison after N days (and N such changes described above.)

Example 1:

Input: cells = [0,1,0,1,1,0,0,1], N = 7
Output: [0,0,1,1,0,0,0,0]
Explanation: 
The following table summarizes the state of the prison on each day:
Day 0: [0, 1, 0, 1, 1, 0, 0, 1]
Day 1: [0, 1, 1, 0, 0, 0, 0, 0]
Day 2: [0, 0, 0, 0, 1, 1, 1, 0]
Day 3: [0, 1, 1, 0, 0, 1, 0, 0]
Day 4: [0, 0, 0, 0, 0, 1, 0, 0]
Day 5: [0, 1, 1, 1, 0, 1, 0, 0]
Day 6: [0, 0, 1, 0, 1, 1, 0, 0]
Day 7: [0, 0, 1, 1, 0, 0, 0, 0]

Example 2:

Input: cells = [1,0,0,1,0,0,1,0], N = 1000000000
Output: [0,0,1,1,1,1,1,0]

8位二进制数每次根据当前值整体进行变化：第i-1和i+1位一样，则i位变为1，否则变为0；问第N次整体变化后结果。因为首尾进行一次变化后会变成0，那一共会有2^6状态。当N>2^6时一定会出现重复，也就是有环，因此用map记录所有出现过的状态并记下出现时的步数t，让N % t 快速跳过所有循环部分。

代码参考自lee251，遍历和t倒过来记录

class Solution:
    def prisonAfterNDays(self, cells: List[int], N: int) -> List[int]:
        m = {}
        while N:
            m[str(cells)] = N
            N -= 1
            cells = [0] + [cells[i - 1] ^ cells[i + 1] ^ 1 for i in range(1, 7)] + [0]
            if str(cells) in m:
                N %= m[str(cells)] - N
        return cells