PrisonCellsAfterNDays.java

// Prison Cells After N Days

// Solution
// 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 {
      public int[] prisonAfterNDays(int[] cells, int N) {
          N = (N - 1) % 14 + 1;
          for (int i = 0; i < N; i++)
              cells = nextDayState(cells);
          return cells;
      }
    
      private int[] nextDayState(int[] cells) {
          int[] next = new int[cells.length];
          for(int i = 1; i < cells.length - 1; i++)
              next[i] = cells[i - 1] == cells[i + 1] ? 1 : 0;
          return next;
       }
    
}