Beautiful Arrangement
<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>
Bit Manipulation; Array; Dynamic Programming; Backtracking; Bitmask
class Solution:
    def countArrangement(self, n: int) -> int:
        def dfs(i):
            nonlocal ans, n
            if i == n + 1:
                ans += 1
                return
            for j in match[i]:
                if not vis[j]:
                    vis[j] = True
                    dfs(i + 1)
                    vis[j] = False

        ans = 0
        vis = [False] * (n + 1)
        match = defaultdict(list)
        for i in range(1, n + 1):
            for j in range(1, n + 1):
                if j % i == 0 or i % j == 0:
                    match[i].append(j)

        dfs(1)
        return ans
