<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 ind, int n, vector<int> &s, map<pair<vector<int>, int>, int> &dp)
    {
        if(ind > n) return 1;

        if(dp[{s, ind}] != 0) return dp[{s, ind}];

        int ans = 0;
        for(int i=1; i<=n; i++)
        {
            if(s[i] == 1 && (i % ind == 0 || ind % i == 0)){
               s[i] = 0;
               ans += solve(ind+1, n, s, dp);
               s[i] = 1;
            }
        } 

        dp[{s, ind}] = ans;
        return ans;
    }
    int countArrangement(int n) {
        vector<int> s(n+1, 1);
        map<pair<vector<int>, int>, int> dp;
        return solve(1, n, s, dp);
    }
};
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