<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:
    bool is_sqr(int n) {
        int s = sqrt(n) + 2;
        while (s * s > n)
            --s;
        return s * s == n;
    }

    int numSquarefulPerms(vector <int> &a) {
        int n = a.size();
        int dp[1 << n][n];
        memset(dp, 0, sizeof(dp));
        for (int i = 0; i < n; i++)
            dp[1 << i][i] = 1;
        vector <int> good[n];
        for (int i = 0; i < n; i++) for (int j = i + 1; j < n; j++)
            if (is_sqr(a[i] + a[j]))
                good[i].push_back(j), good[j].push_back(i);
        for (int mask = 1; mask < (1 << n); ++mask) {
            for (int i = 0; i < n; i++) if (!(mask >> i & 1)) {
                for (auto &w : good[i]) if (mask >> w & 1) {
                    dp[mask | (1 << i)][i] += dp[mask][w];
                }
            }
        }
        int ans = 0;
        for (int last = 0; last < n; last++)
            ans += dp[(1 << n) - 1][last];
        sort(a.begin(), a.end());
        for (int i = 0; i < n; i++) {
            int j = i;
            while (j < n - 1 and a[j + 1] == a[j])
                ++j, ans /= j - i + 1;
            i = j;
        }
        return ans;
    }
};
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