<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 dp[1<<12][13];
    int numSquarefulPerms(vector<int>& nums) {
        memset(dp,-1,sizeof(dp));
        int n=nums.size();
        return find(nums,0,n,n);
    }
    int find(vector<int>&nums,int mask,int p,int &n){
        if(mask==(1<<n)-1)return 1;
        if(dp[mask][p]!=-1)return dp[mask][p];
        int ans=0;
        map<int,int>mp;
        for(int i=0;i<n;i++){
            if(mask&(1<<i) || mp[nums[i]]>0)continue;
            mp[nums[i]]++;
            if(p==n){
                ans+=find(nums,mask|(1<<i),i,n);
            }
            else{
                int x=nums[p]+nums[i];
                int y=sqrt(x);
                if(x==y*y )ans+=find(nums,mask|(1<<i),i,n);
            }
        }
        return dp[mask][p]=ans;
    }
};
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