Suppose you have N integers from 1 to N. We define a beautiful arrangement as an array that is constructed by these N numbers successfully if one of the following is true for the ith position (1 <= i <= N) in this array:

1. The number at the ith position is divisible by i.
2. i is divisible by the number at the ith position.

 

Now given N, how many beautiful arrangements can you construct?

Example 1:

Input: 2
Output: 2
Explanation: 

The first beautiful arrangement is [1, 2]:

Number at the 1st position (i=1) is 1, and 1 is divisible by i (i=1).

Number at the 2nd position (i=2) is 2, and 2 is divisible by i (i=2).

The second beautiful arrangement is [2, 1]:

Number at the 1st position (i=1) is 2, and 2 is divisible by i (i=1).

Number at the 2nd position (i=2) is 1, and i (i=2) is divisible by 1.

 

Note:

1. N is a positive integer and will not exceed 15.

 

Approach #1: backtracking. [C++]

class Solution {
public:
    int countArrangement(int N) {
        vector<int> path;
        
        for (int i = 1; i <= N; ++i) 
            path.push_back(i);
        
        return helper(N, path);
    }
    
private:
    int helper(int n, vector<int> path) {
        if (n <= 0) return 1;
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            if (path[i] % n == 0 || n % path[i] == 0) {
                swap(path[i], path[n-1]);
                ans += helper(n-1, path);
                swap(path[i], path[n-1]);
            }
        }
        return ans;
    }
};