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.

You are given an integer array cells where cells[i] == 1 if the ith cell is occupied and cells[i] == 0 if the ith cell is vacant, and you are given an integer n.

Return the state of the prison after n days (i.e., 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]

 

Constraints:

  • cells.length == 8
  • cells[i] is either 0 or 1.
  • 1 <= n <= 109

submission

// successors trick
// since the number of possible state is at most 2^6
// there is a loop, and others brute forced the 
// loop pattern to be of length 1, 7 or 14
// another observation: every state must be in loop
// hence we just simulate (n - 1) % 14 + 1 steps
impl Solution {
    pub fn prison_after_n_days(cells: Vec<i32>, n: i32) -> Vec<i32> {
        std::iter::successors(Some(cells), |prev| Some(
            (0..8).map(|idx| match idx {
                0 | 7 => 0,
                idx => (prev[idx - 1] == prev[idx + 1]) as i32,
            })
            .collect()
        ))
        // note with successors we need +1 here
        .take((n as usize - 1) % 14 + 2)
        .last()
        .unwrap()
    }
}