<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 n = nums.size();
        vector<vector<int>> adj(n);
        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];
                int root = sqrt(sum);
                if (root * root == sum) {
                    adj[i].push_back(j);
                }
            }
        }

        vector<bool> used(n, false);
        int res = 0;

        // Start DFS from each unique number
        for (int i = 0; i < n; ++i) {
            if (i > 0 && nums[i] == nums[i - 1]) continue; // Skip duplicates
            used[i] = true;
            dfs(nums, adj, used, i, 1, res);
            used[i] = false;
        }

        return res;
    }

private:
    void dfs(const vector<int>& nums, const vector<vector<int>>& adj, vector<bool>& used, int idx, int count, int& res) {
        if (count == nums.size()) {
            ++res;
            return;
        }

        unordered_set<int> seen; // To avoid duplicates at the same tree level
        for (int next : adj[idx]) {
            if (used[next]) continue;
            if (seen.count(nums[next])) continue; // Skip duplicates
            used[next] = true;
            seen.insert(nums[next]);
            dfs(nums, adj, used, next, count + 1, res);
            used[next] = false;
        }
    }
};

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