<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:
    vector<int> nums;
    int dp[13][13][4097];

    bool check(int a, int b){
          int s = sqrt(a+b);

          return (s*s==(a+b));
    }

    int rec(int lev, int prev, int mask){
        if(lev==-1)return 1;
        if(dp[lev][prev][mask]!=-1)return dp[lev][prev][mask];

        int ans = 0;
        for(int i=0;i<nums.size();i++){
             if(!(mask&(1<<i))){
                if(i>0 && nums[i]==nums[i-1]&& (mask&(1<<(i-1))))continue;
                if(check(nums[i],nums[prev]))ans+=rec(lev-1,i,mask|(1<<i));
             }
        }

        return dp[lev][prev][mask]=ans;
    }
    int numSquarefulPerms(vector<int>& nm) {
        nums=nm;
        sort(nums.begin(), nums.end());
        memset(dp,-1,sizeof(dp));
        int ans=0;

        for(int i=0;i<nm.size();i++){
            ans+=rec(nm.size()-2,i,0|(1<<i));
        }

    
        return ans;
    }
};
12400