StarWalker

Where my heart will take me

0%

Leetcode957-prisonCellsAfterNDays

Posted on Edited on In Views: Valine:
Solution Report of LeetCode Acceptted

Description

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

Example 1:

1
2
3
4
5
6
7
8
9
10
11
12
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:
1
2
Input: cells = [1,0,0,1,0,0,1,0], N = 1000000000
Output: [0,0,1,1,1,1,1,0]

Solution

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
class Solution {
    public int[] prisonAfterNDays(int[] cells, int N) {
        Map<String, Integer> seen = new HashMap<>();
        while (N > 0) {
            int[] cells2 = new int[8];
            seen.put(Arrays.toString(cells), N--);
            for (int i = 1; i < 7; ++i)
                cells2[i] = cells[i - 1] == cells[i + 1] ? 1 : 0;
            cells = cells2;
            if (seen.containsKey(Arrays.toString(cells))) {
                N %= seen.get(Arrays.toString(cells)) - N;
            }
        }
        return cells;
    }
}