<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:
    def numSquarefulPerms(self, nums: List[int]) -> int:
        def is_perfect_square(x):
            s = int(math.sqrt(x))
            return s*s ==x
        
        nums.sort()
        n = len(nums)
        def backtrack(path, used):
            if len(path) == n:
                nonlocal count
                count += 1
                return
            
            for i in range(n):
                if i in used or (i > 0 and nums[i] == nums[i-1]  and (i-1) not in used):
                    continue
                if len(path) > 0 and not is_perfect_square(path[-1] + nums[i]):
                    continue
                used.add(i)
                path.append(nums[i])
                backtrack(path, used)
                used.remove(i)
                path.pop()
        count = 0
        used = set()
        backtrack([], used)
        return count


        
16500