<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 check(int a,int b){
        int temp = sqrt(a+b);
        return ((temp*temp) == (a+b));
    }

int helper(int &visit,vector<int>&ds,int n,int &ans,vector<int>& nums){
    if(ds.size()==n){
        return 1;
    }
    int ways=0;
    for(int i=0;i<n;i++){
        int k=i-1;
        if ((visit & (1 << i)) != 0 || (i > 0 && nums[i] == nums[i - 1] && (visit & (1 << (i - 1))) == 0))continue;
        
        if(ds.size()==0 || check(ds.back(),nums[i])){
            visit |= (1 << i);
            ds.push_back(nums[i]);
            ways+=helper(visit,ds,n,ans,nums);
            ds.pop_back();
            visit ^= (1 << i);
        }
        
    }
    return ways;
}

int numSquarefulPerms(vector<int>& nums) {
    int n=nums.size();
    sort(nums.begin(),nums.end());
    int ans=0,visit=0;
    vector<int>ds;
    return helper(visit,ds,n,ans,nums);;
}
};
9200