<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:
    bool checkSquare(long long num){
        long long root=sqrt(num);
        return (root*root)==num;
    }
    int dp[1<<13][13];
    int memo(int mask,int lastIdx,vector<int>&nums){
        if(mask==(1<<nums.size())-1) return 1;
        int &res=dp[mask][lastIdx];
        if(~res) return res;
        res=0;
        for(int i=0;i<nums.size();i++){
            if(mask&(1<<i)) continue;
            if(i>0&&nums[i]==nums[i-1]&&!(mask&(1<<(i-1)))) continue;
            if(lastIdx==12||checkSquare(nums[i]+nums[lastIdx])){
                res+=memo(mask|(1<<i),i,nums);
            }
        }
        return res;
    }
    int numSquarefulPerms(vector<int>& nums) {
        memset(dp,-1,sizeof(dp));
        sort(nums.begin(),nums.end());
        return memo(0,12,nums);
    }
};
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