<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 {
    int dp[13][13][(1<<12)];
    int n;
    vector<int> v;
public:
    bool check(long long int num){
        long long int low=0;
        long long int high=1e9;
        long long int ans=0;
        while(low<=high){
            long long int mid=(low+high)/2;
            if(mid*mid<=num){
                ans=mid;
                low=mid+1;
            }
            else{
                high=mid-1;
            }
        }        
        return ans*ans==num;
    }
    int rec(int prev,int level,int mask){
        // cout<<prev<<" "<<level<<" "<<mask<<endl;
        if(level==n){
            return 1;
        }
        if(dp[prev][level][mask]!=-1){
            return dp[prev][level][mask];
        }
        int ans=0;
        unordered_map<int,int> mp;
        for(int i=0;i<n;i++){
            if(!(mask&(1<<i))&&check(v[prev]+v[i])){
                if(!mp[v[i]]){
                    ans+=rec(i,level+1,mask|(1<<i));
                    mp[v[i]]=1;
                }
            }
        }
        return dp[prev][level][mask]=ans;
    }
    int numSquarefulPerms(vector<int>& nums) {
        memset(dp,-1,sizeof(dp));
        v=nums;
        n=v.size();
        int ans=0;
        unordered_map<int,int> mp;
        for(int i=0;i<n;i++){
            if(!mp[v[i]]){
                ans+=rec(i,1,1<<i);
                mp[v[i]]=1;
            }
        }
        return ans;
    }
};
13000