# 957. 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 neighbours 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 neighbours. you are given an integer array 'cells' where 'cells[i] == 1' if the i'th cell is occupied and 'cells[i] == 0' if the i'th cell is vacant, and you are given an integer 'n'. return the state of the prison after 'n' days (ie. 'n' such changes desribed above) """ class Solution(object): def prisonAfterNDays(self, cells, n): """ :type cells: List[int] :type n: int :rtype: List[int] """ vis = {str(cells): n} while n: vis.setdefault(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 vis: n %= vis[str(cells)] - n return cells if __name__ == "__main__": obj = Solution() print(obj.prisonAfterNDays(cells=[0, 1, 0, 1, 1, 0, 0, 1], n=7)) print(obj.prisonAfterNDays(cells=[1, 0, 0, 1, 0, 0, 1, 0], n=1000000000))