<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 {
public:
bool isPerfectSquare(long long x)
{
    if (x >= 0) {
        long long sr = sqrt(x);
        return (sr * sr == x);
    }
    return false;
}
    int solve(vector<int>&nums,int ind)
    {
        if(ind==nums.size()) return 1;
        int ans=0;
        for(int i=ind;i<nums.size();i++)
        {
            if(i>ind&&(nums[i]==nums[i-1]||nums[i]==nums[ind])) continue;
            swap(nums[i],nums[ind]);
            if(ind==0||isPerfectSquare(nums[ind]+nums[ind-1])) {
                ans=ans+solve(nums,ind+1);
            }
            swap(nums[i],nums[ind]);
        }
        return ans;
    }
    int numSquarefulPerms(vector<int>& nums) {
        sort(nums.begin(),nums.end());
        return solve(nums,0);
    }
};
9100