<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;
    unordered_map<int, int>map1;
    unordered_map<int, unordered_set<int>>map2;

    void dfs(int i, int left) {
        map1[i]--;
        if (!left) ans++;
        for (auto j : map2[i]) {
            if (map1[j] > 0) {
                dfs(j, left - 1);
            }
        }
        map1[i]++;
    }

    int numSquarefulPerms(vector<int>& nums) {
        for (int i : nums) map1[i]++;
        for (auto i : map1) {
            for (auto j : map1) {
                int x = i.first, y = j.first;
                int s = sqrtl(x + y);
                if (s * s == x + y) {
                    map2[x].insert(y);
                }
            }
        }

        for (auto i : map1) {
            dfs(i.first, nums.size() - 1);
        }
        return ans;
    }

};
10000