<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 {
public:
    int solve(int i, int n, unordered_set<int> &us){
        // base case
        if (i==0) return ((int)us.size()==n-1);

        int ans=0;
        // choices 1, 2, 3, 4, 5 ... n
        for(int j=1; j<n+1; j++){
            if (us.find(j)!=us.end()) continue;
            if ((i+1)%(j) == 0 or (j)%(i+1) == 0) {
                // use j
                us.insert(j);
                ans += solve(i-1, n, us);
                us.erase(j);
            }
        }
        return ans;
    }
    int countArrangement(int n) {
        // 1 3 2
        // 1 2 3
        unordered_set<int> us;
        int ans = solve(n-1, n, us);
        return ans;
    }
};
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