<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:
long long solve(vector<int>& nums,int mask,vector<vector<long long>> &dp,int prev)
{
    int n=nums.size();
    if(mask==0)return 1;
    if(prev==-1)
    {
        long long ans=0;
            for(int i=0;i<n;i++)
            {
                int m=(1<<i);
                    ans+=solve(nums,(mask^m),dp,i);
            }
            return ans;
    }
    if(dp[prev][mask]!=-1)return dp[prev][mask];
    long long ans=0;
    for(int i=0;i<n;i++)
    {
        int m=(1<<i);
        if((m&mask)!=0)
        {
            int x=nums[i]+nums[prev];
            double d=(double)sqrt(x);
            if(floor(d)==ceil(d))
            {
                ans+=solve(nums,(mask^m),dp,i);
            }
        }
        
    }
    
    dp[prev][mask]=ans;
    return ans;
}
    int numSquarefulPerms(vector<int>& nums) {
        int n=nums.size();
        vector<vector<long long>> dp(n,vector<long long>((1<<12),-1));
        long long ans=solve(nums,(1<<n)-1,dp,-1);
        map<int,int> mp;
        long long fact[13];
        fact[0]=1;
        for(int i=1;i<=12;i++)
        {
            fact[i]=(long long)i*fact[i-1];
        }
        for(int i=0;i<n;i++)
        {
            mp[nums[i]]++;
        }
        for(auto i=mp.begin();i!=mp.end();i++)
        {
            ans=ans/fact[mp[i->first]];
        }
        return ans;
    }
};
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