Number of Squareful Arrays
<p>An array is <strong>squareful</strong> if the sum of every pair of adjacent elements is a <strong>perfect square</strong>.</p>

<p>Given an integer array nums, return <em>the number of permutations of </em><code>nums</code><em> that are <strong>squareful</strong></em>.</p>

<p>Two permutations <code>perm1</code> and <code>perm2</code> are different if there is some index <code>i</code> such that <code>perm1[i] != perm2[i]</code>.</p>

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

<pre>
<strong>Input:</strong> nums = [1,17,8]
<strong>Output:</strong> 2
<strong>Explanation:</strong> [1,8,17] and [17,8,1] are the valid permutations.
</pre>

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

<pre>
<strong>Input:</strong> nums = [2,2,2]
<strong>Output:</strong> 1
</pre>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 12</code></li>
	<li><code>0 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
</ul>
Bit Manipulation; Array; Hash Table; Math; Dynamic Programming; Backtracking; Bitmask
class Solution:
    def numSquarefulPerms(self, nums: List[int]) -> int:
        n = len(nums)
        f = [[0] * n for _ in range(1 << n)]
        for j in range(n):
            f[1 << j][j] = 1
        for i in range(1 << n):
            for j in range(n):
                if i >> j & 1:
                    for k in range(n):
                        if (i >> k & 1) and k != j:
                            s = nums[j] + nums[k]
                            t = int(sqrt(s))
                            if t * t == s:
                                f[i][j] += f[i ^ (1 << j)][k]

        ans = sum(f[(1 << n) - 1][j] for j in range(n))
        for v in Counter(nums).values():
            ans //= factorial(v)
        return ans
