"""Leetcode 957. Prison Cells After N Days
Medium

URL: 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 SolutionPattern(object):
    def prisonAfterNDays(self, cells, N):
        """
        :type cells: List[int]
        :type N: int
        :rtype: List[int]
        """
        # Since the cells transformation will cycle after 14 days.
        N %= 14
        if N == 0:
            N = 14

        for _ in range(N):
            res = [0] * len(cells)

            for i in range(1, len(cells) - 1):
                if cells[i - 1] == cells[i + 1]:
                    res[i] = 1

            cells = res

        return res


def main():
    # Output: [0, 0, 1, 1, 0, 0, 0, 0]
    cells = [0,1,0,1,1,0,0,1]
    N = 7
    print SolutionPattern().prisonAfterNDays(cells, N)

    # Output: [0,0,1,1,1,1,1,0]
    cells = [1,0,0,1,0,0,1,0]
    N = 1000000000
    print SolutionPattern().prisonAfterNDays(cells, N)


if __name__ == '__main__':
    main()

Complete finding pattern sol