<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 {
    bool vis[13];
    vector<int> g[13];
public:
    int numSquarefulPerms(vector<int>& nums) {
        int n = nums.size();
        sort(nums.begin(), nums.end());
        for(int i=0; i<n; i++){
            for(int j=0; j<n; j++){
                if(i==j) continue;

                int sum = nums[i]+nums[j];
                if(sqrt(sum)==(int)sqrt(sum)){
                    g[i].push_back(j);
                }
            }
        }
        int cnt = 0;
        int prev = -1;
        for(int i=0; i<n; i++){
            if(nums[i]==prev) continue;
            vis[i] = true;
            cnt += helper(nums, i, 1);
            vis[i] = false;
            prev = nums[i];
        }
        return cnt;
    }

    int helper(const vector<int>& nums, int cur, int cnt){
        if(cnt>=nums.size()) return 1;

        int res = 0;
        int prev = -1;
        for(auto neighbor: g[cur]){
            if(vis[neighbor]) continue;
            
            if(nums[neighbor]==prev) continue;

            vis[neighbor] = true;
            res += helper(nums, neighbor, cnt+1);
            vis[neighbor] = false;
            prev = nums[neighbor];
        }
        return res;
    }
};
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