<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:
    int t[13][1<<13][13];
    bool check(long long int x){
        long long int y= sqrt(x);
        return (y*y==x);
    }
    int f(int i, int mask, int prev, vector<int>& nums){
         int n= nums.size();
        if(i==n){
            return 1;
        }
        if(t[i][mask][prev+1]!=-1)return t[i][mask][prev+1];
        int ans=0;
        for(int j=0; j<n; j++){
            
            if((mask& (1<<j))==0){
                if(j>0 && nums[j]==nums[j-1] && !(mask&(1<<(j-1)))){
                   continue;
                } 
                if(prev==-1 || check(nums[j]+nums[prev])==true){
                  
                    ans+=f(i+1, mask|(1<<j),j,nums);
                  
                }
               
            }
          

        }
        return t[i][mask][prev+1]=ans;
    }
    int numSquarefulPerms(vector<int>& nums) {
         int n= nums.size();
         memset(t,-1,sizeof(t));
         sort(nums.begin(), nums.end());
         return f(0,0,-1,nums);
    }
};
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