Problem

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

Pre analysis

Will simulate the steps

Post analysis

Simulation didn’t work for very huge N, (eg, 100k), As there are 8 binary fields, there are at most 2^8 = 256 codewords we can construct. If we were to start from the first cell and simulate this we would eventually hit a codeword we already have. In effect, this makes it 6 cells, each one could be occupied or not, that gives us 6C2 = 15. Since the first and last cell don’t change and will always be 0 regardless of their initial value, that means that the 2nd and 7th cell will flip their state if and only if the 3rd and 6th cells are zero, so that’s a combination that will never happen (2nd and 7th cell with value 1 and 3rd and 6th cell with value 1) which gives us in total 6C2 - 1 = 14 combination. After 14 days, the combination will just repeat over and over again, which leads to the following optimized constant time solution.