<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 ans = 0, n;
    unordered_set<int> squares;
public:
    int numSquarefulPerms(vector<int>& nums) {
        ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
        n = nums.size();
        sort(nums.begin(), nums.end());

        for(int i = 0; i * i <= *max_element(nums.begin(), nums.end()) * 2; ++i)
            squares.insert(i * i);

        backtrack(nums, 0);
        return ans;
    }
    void backtrack(vector<int> nums, int pos){
        if(pos == n)
            ++ans;
        
        for(int i = pos; i < n; ++i){
            if(i > pos && nums[pos] == nums[i])
                continue;
            swap(nums[pos], nums[i]);
            if(!pos || (pos && squares.contains(nums[pos] + nums[pos - 1])))
                backtrack(nums, pos + 1);
        }
    }

};
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