<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 numSquarefulPerms(vector<int>& nums) {
        int ans = 0;
        vector<int> cur;
        vector<int> used(nums.size(), 0);
        sort(nums.begin(), nums.end());
        dfs(nums, used, cur, ans);
        return ans;
    }
private:
    bool isSquareful(int x, int y) {
        int s = sqrt(x + y);
        return s * s == x + y;
    }
    
    void dfs(const vector<int>& nums, vector<int>& used, vector<int>& cur, int& ans) {
        if (cur.size() == nums.size()) {
            ans++;
            return;
        }
        for (int i = 0; i < nums.size(); i++) {
            if (used[i]) continue;
            if (i > 0 && nums[i] == nums[i - 1] && !used[i - 1]) continue;
            if (!cur.empty() && !isSquareful(cur.back(), nums[i])) continue;

            used[i] = 1;
            cur.push_back(nums[i]);
            dfs(nums, used, cur, ans);
            used[i] = 0;
            cur.pop_back();
        }
    }
};
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