Question
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
Difficulty:Medium
Category:
Analyze
Solution
class Solution {
public:
vector<int> prisonAfterNDays(vector<int>& cells, int N) {
N = N % 14;
if (N == 0) N = 14;
for (int i = 1; i <= N; ++i) {
vector<int> temp = cells;
for (int j = 0; j < temp.size(); ++j) {
if (j == 0 || j == temp.size() - 1) {
cells[j] = 0;
continue;
} else {
if (temp[j - 1] == temp[j + 1]) {
cells[j] = 1;
} else {
cells[j] = 0;
}
}
}
}
return cells;
}
};