Number of Squareful Arrays
<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>
Bit Manipulation; Array; Hash Table; Math; Dynamic Programming; Backtracking; Bitmask
class Solution {
public:
    int numSquarefulPerms(vector<int>& nums) {
        int n = nums.size();
        int f[1 << n][n];
        memset(f, 0, sizeof(f));
        for (int j = 0; j < n; ++j) {
            f[1 << j][j] = 1;
        }
        for (int i = 0; i < 1 << n; ++i) {
            for (int j = 0; j < n; ++j) {
                if ((i >> j & 1) == 1) {
                    for (int k = 0; k < n; ++k) {
                        if ((i >> k & 1) == 1 && k != j) {
                            int s = nums[j] + nums[k];
                            int t = sqrt(s);
                            if (t * t == s) {
                                f[i][j] += f[i ^ (1 << j)][k];
                            }
                        }
                    }
                }
            }
        }
        long long ans = 0;
        for (int j = 0; j < n; ++j) {
            ans += f[(1 << n) - 1][j];
        }
        unordered_map<int, int> cnt;
        for (int x : nums) {
            ++cnt[x];
        }
        int g[13] = {1};
        for (int i = 1; i < 13; ++i) {
            g[i] = g[i - 1] * i;
        }
        for (auto& [_, v] : cnt) {
            ans /= g[v];
        }
        return ans;
    }
};