<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 {
private:
    int nFactorial(int n){
        if(n==1) return 1;
        return n * nFactorial(n-1);
    }
public:
    int numSquarefulPerms(vector<int>& nums) {
        map<int,int> cnt;
        int n = nums.size();
        vector<vector<int>> match(n);
        for(int i=0; i<n; ++i){
            cnt[nums[i]]++;
            for(int j=i+1; j<n; ++j){
                int num = nums[i]+nums[j];
                for(int p=0; p*p<=num; ++p){
                    if(p*p==num) {
                        match[i].push_back(j);
                        match[j].push_back(i);
                        break;
                    }
                }
            }
        }
        vector<unordered_map<int,int>> dp(1<<n);
        for(int i=0; i<n; ++i){
            dp[1<<i][i] = 1;
        }
        for(int i=1; i<(1<<n); ++i){
            for(auto& x:dp[i]){
                for(int next:match[x.first]){
                    if(i & (1<<next)) continue;
                    dp[i | (1<<next)][next] += x.second;
                }
            }
        }
        int ans = 0;
        for(auto x:dp[(1<<n)-1]) ans += x.second;
        for(auto x:cnt){
            ans /= nFactorial(x.second);
        }
        return ans;
    }
};

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