<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:
    map<int,int> count;
    map<int,unordered_set<int>> mp;
    int answer;
    int numSquarefulPerms(vector<int>& nums) {
        for(auto &it : nums)
        count[it]++;
        for(auto &a : count)
        {
            for(auto &b: count)
            {
                int x = sqrt(a.first+b.first);
                if(x*x==a.first+b.first)
                {
                    mp[a.first].insert(b.first);
                }
            }
        }

        for(auto it : count)
        {
            dfs(it.first,nums.size()-1);
        }

        return answer;
    }

    void dfs(int it , int n)
    {
        count[it]--;
        if(!n) answer++;
        for(auto a : mp[it])
        {
            if(count[a]>0)
            dfs(a,n-1);
        }

        count[it]++;
    }
};
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