Prison Cells After N Days

Similar Problems:

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

Github: code.dennyzhang.com

Credits To: leetcode.com

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