0957.Prison_Cells_After_N_Days.py

"""
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.

You are given an integer array cells where cells[i] == 1 if the ith cell is occupied and cells[i] == 0 if the ith cell is vacant, and you are given an integer n.

Return the state of the prison after n days (i.e., 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]
 

Constraints:

cells.length == 8
cells[i] is either 0 or 1.
1 <= n <= 109
"""

#refer: https://maxming0.github.io/2020/07/03/Prison-Cells-After-N-Days/
#map(str,cells) 
class Solution:
    def prisonAfterNDays(self, cells: List[int], n: int) -> List[int]:
        dic = {}
        while n > 0:
            dic[''.join(map(str,cells))] = n     #将int转成string
            n -= 1
            temp = [0] * 8
            for i in range(1, 7):
                temp[i] = 1 if cells[i - 1] == cells[i + 1] else 0
            cells = temp
            t = ''.join(map(str, cells))
            if t in dic:
                n = n % (dic[t] - n)
        return cells