957-prison-cells-after-n-days.py

"""
Problem Link: 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]
 
Note:
cells.length == 8
cells[i] is in {0, 1}
1 <= N <= 10^9
"""
class Solution:
    def next_step(self, cells):
        res = [0] * 8
        for i in range(1,7):
            res[i] = int(cells[i-1] == cells[i+1])
        return res
    
    def prisonAfterNDays(self, cells: List[int], N: int) -> List[int]:
            
        found_dic = {}
        for i in range(N):
            cells_str = str(cells)
            if cells_str in found_dic:
                loop_len = i - found_dic[cells_str]
                return self.prisonAfterNDays(cells, (N - i) % loop_len)
            else:
                found_dic[cells_str] = i 
                cells = self.next_step(cells) 
                
        return cells