<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:
    ans = 0
    used = [False] * len(nums)

    def isSquare(num: int) -> bool:
      root = int(math.sqrt(num))
      return root * root == num

    def dfs(path: List[int]) -> None:
      nonlocal ans
      if len(path) > 1 and not isSquare(path[-1] + path[-2]):
        return
      if len(path) == len(nums):
        ans += 1
        return

      for i, a in enumerate(nums):
        if used[i]:
          continue
        if i > 0 and nums[i] == nums[i - 1] and not used[i - 1]:
          continue
        used[i] = True
        dfs(path + [a])
        used[i] = False

    nums.sort()
    dfs([])
    return ans
16700