<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>

class Solution:
    @functools.lru_cache()
    def is_squareful(self, n):
        root = int(math.sqrt(n))
        return root * root == n

    def numSquarefulPerms(self, nums: List[int]) -> int:
        s = set(nums)
        next_val_map = defaultdict(set)
        for n1 in s:
            for n2 in s:
                if self.is_squareful(n1 + n2):
                    next_val_map[n1].add(n2)

        counter = defaultdict(int)
        for n in nums:
            counter[n] += 1
        res = 0

        def dfs():
            nonlocal res

            if len(existing_list) == len(nums):
                res += 1
                return
            for next_val in next_val_map[existing_list[-1]]:
                if counter[next_val] > 0:
                    existing_list.append(next_val)
                    counter[next_val] -= 1
                    dfs()
                    existing_list.pop()
                    counter[next_val] += 1

        for n in s:
            existing_list = [n]
            counter[n] -= 1
            dfs()
            counter[n] += 1            

        return res

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