<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:
    vector<unordered_map<int, int>> dp;
    vector<int> pre;
    int func(int mask, int n, vector<int>& nums, int prev) {
        int i = __builtin_popcount(mask);
        if (i == n)
            return 1;
        if (prev != -1 && dp[mask].find(nums[prev]) != dp[mask].end())
            return dp[mask][nums[prev]];
        int ans = 0;
        for (int j = 0; j < n; j++) {
            bool ok = true;
            int sum = nums[j];
            if (prev != -1) {
                sum += nums[prev];
            }
            int sqr = sqrt(sum);
            if (prev != -1)
                ok = (sqr * sqr == sum);
            if (!(mask & (1 << j)) && ok) {
                ans = (ans + func(mask | (1 << j), n, nums, j));
            }
        }
        if (prev != -1)
            dp[mask][nums[prev]] = ans;
        return ans;
    }
    int numSquarefulPerms(vector<int>& nums) {
        int n = nums.size();
        dp.resize(1 << n);
        pre.resize(13);
        pre[0] = 1;
        for (int i = 1; i <= 12; i++) {
            pre[i] = pre[i - 1] * i;
        }
        unordered_map<int, int> mp;
        for (int i = 0; i < n; i++) {
            mp[nums[i]]++;
        }
        int ans = func(0, n, nums, -1);
        for (auto it : mp) {
            ans = (ans / pre[it.second]);
        }
        return ans;
    }
};
18800