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]
Brute Force -> LTE
class Solution { public int[] prisonAfterNDays(int[] cells, int N) { int[][] arr = new int[N + 1][cells.length]; for (int i = 0; i < cells.length; i++) { arr[0][i] = cells[i]; } for (int i = 1; i <= N; i++) { for (int j = 0; j < cells.length; j++) { if (j == 0 || j == cells.length - 1) { continue; } if (arr[i - 1][j - 1] == 0 && arr[i - 1][j + 1] == 0 || arr[i - 1][j - 1] == 1 && arr[i - 1][j + 1] == 1) { arr[i][j] = 1; } } // System.out.println(i + ": " + Arrays.toString(arr[i])); } return arr[N]; } }
solution: find there is cycle form
class Solution { public int[] prisonAfterNDays(int[] cells, int N) { Set<String> set = new HashSet<>(); boolean hasCycle = false; int count = 0; int[] nxt = new int[cells.length]; for (int i = 0; i < N; i++) { nxt = getNext(cells); String str = Arrays.toString(nxt); if (set.contains(str)) { hasCycle = true; break; } else { set.add(str); count += 1; } cells = nxt; } if (hasCycle) { N = N % count; for (int i = 0; i < N; i++) { cells = getNext(cells); } } return cells; } private int[] getNext(int[] cells) { int[] arr = new int[cells.length]; for (int i = 1; i < cells.length - 1; i++) { arr[i] = cells[i - 1] == cells[i + 1] ? 1 : 0; } return arr; } }