<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>

from math import sqrt, factorial
from typing import List

class Solution:
    def numSquarefulPerms(self, nums: List[int]) -> int:
        count = {k: nums.count(k) for k in nums}

        redundancy = 1
        for v in count.values():
            redundancy *= factorial(v)

        nums.sort()

        def isSquare(n):
            x = int(sqrt(n))
            return x * x == n

        n = len(nums)
        dp = [[0] * (1 << n) for _ in range(n + 1)]
        dp[n][0] = 1

        for mask in range(1 << n):
            for i in range(n):
                if mask & (1 << i):
                    for j in range(n):
                        if mask & (1 << j) and isSquare(nums[i] + nums[j]):
                            dp[i][mask] += dp[j][mask ^ (1 << i)]
                elif mask == 0:
                    dp[i][mask | (1 << i)] = 1

        res = 0
        for i in range(n):
            res += dp[i][(1 << n) - 1]

        return res // redundancy

17200