<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:
    using ll = long long;
    vector<vector<ll>> dp;
    ll fact(int x) {
        ll ans = 1;
        while (x > 0) {
            ans *= x;
            x--;
        }
        return ans;
    }
    int numSquarefulPerms(vector<int>& nums) {
        int n = nums.size();
        dp = vector<vector<ll>>(1 << n, vector<ll>(n, -1));
        ll ans = 0;
        for (int i = 0; i < n; i++) {
            ans += solve(1 << i, i, nums);
        }
        unordered_map<int, int> cnt;
        for (auto v : nums) {
            cnt[v]++;
        }
        for (const auto& [_, v] : cnt) {
            ans /= fact(v);
        }
        return ans;
    }

    bool isS(ll v) {
        ll a = 0, b = 1e9;
        while (a <= b) {
            ll m = (a + b) >> 1;
            ll x = m * m;
            if (x < v) {
                a = m + 1;
            } else if (x > v) {
                b = m - 1;
            } else {
                return true;
            }
        }
        return false;
    }

    ll solve(int state, ll last, const vector<int>& nums) {
        int n = nums.size();
        if (state == ((1 << n) - 1)) {
            return 1;
        }
        if (dp[state][last] != -1) {
            return dp[state][last];
        }
        dp[state][last] = 0;
        for (int i = 0; i < n; i++) {
            if (!(state & (1 << i))) {
                int nxt = (state | (1 << i));
                if (isS((ll)(nums[i]) + nums[last])) {
                    dp[state][last] += solve(nxt, i, nums);
                }
            }
        }
        return dp[state][last];
    }
};
19000