<p>Suppose you have <code>n</code> integers labeled <code>1</code> through <code>n</code>. A permutation of those <code>n</code> integers <code>perm</code> (<strong>1-indexed</strong>) is considered a <strong>beautiful arrangement</strong> if for every <code>i</code> (<code>1 &lt;= i &lt;= n</code>), <strong>either</strong> of the following is true:</p>

<ul>
	<li><code>perm[i]</code> is divisible by <code>i</code>.</li>
	<li><code>i</code> is divisible by <code>perm[i]</code>.</li>
</ul>

<p>Given an integer <code>n</code>, return <em>the <strong>number</strong> of the <strong>beautiful arrangements</strong> that you can construct</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> n = 2
<strong>Output:</strong> 2
<b>Explanation:</b> 
The first beautiful arrangement is [1,2]:
    - perm[1] = 1 is divisible by i = 1
    - perm[2] = 2 is divisible by i = 2
The second beautiful arrangement is [2,1]:
    - perm[1] = 2 is divisible by i = 1
    - i = 2 is divisible by perm[2] = 1
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = 1
<strong>Output:</strong> 1
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 15</code></li>
</ul>

class Solution {
private:
    void generate_perm(int n, int mask, vector<int>& v, int ct, vector<vector<int>>& perms){
        if(ct==n){
            perms.push_back(v);
            return;
        }

        for(int i=1;i<=n;i++){
            int curr_mask=1<<(i-1);
            if(!(mask&curr_mask)){
                if(i%(ct+1)==0 || (ct+1)%i==0){
                    v.push_back(i);
                    generate_perm(n,(mask|curr_mask),v,ct+1,perms);
                    v.pop_back();
                }
            }
        }

        return;
    }
public:
    int countArrangement(int n) {
        vector<int> v;
        vector<vector<int>> perms;
        int ct=0,mask=0;
        generate_perm(n,mask,v,ct,perms);

        int ans=0;
        for(auto it : perms){
            int flag=1;
            for(int i=0;i<n;i++){
                if(!(it[i]%(i+1)==0 || (i+1)%it[i]==0)){
                    flag=0;
                    break;
                }
            }

            if(flag)ans++;
        }

        return ans;
    }
};
20997