<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:
    int ans = 0;
    bool check(int n){
        int root = sqrt(n);
        return ((root*root) == n);
    }

    void help(int ind, vector<int> nums){ // dont take by reference as we use swapping way of permutations 
        if(ind >= nums.size()){
            ans++;
            return;
        }

        for(int i = ind; i < nums.size();i++){
            if(i == ind || nums[i] != nums[ind]){
                swap(nums[i], nums[ind]);
                if(ind == 0 || (ind > 0 && check(nums[ind-1] + nums[ind]))){
                    help(ind+1,nums);
                }
            }
        }
        return;
    }
    int numSquarefulPerms(vector<int>& nums) {
        int n = nums.size();
        sort(nums.begin(),nums.end());
        help(0,nums);
        return ans;
    }
};
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