Brian Yang
Brian Yang
algorithm designer
Mar 9, 2020 2 min read

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:

    1. cells.length == 8
    2. cells[i] is in {0, 1}
    3. 1 <= N <= 10^9

    Solution

    var prisonAfterNDays = function (cells, N) {
  let copy = cells.slice()
  let maxIter = 2 * cells.length - 2
  N = N % maxIter === 0 ? maxIter : N % maxIter
  while (N-- > 0) {
    for (let i = 0; i < cells.length; i++) {
      copy[i] = (cells[i - 1] === cells[i + 1]) ? 1 : 0
    }
    cells = copy.slice()
  }
  return cells;
};
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