<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 {

    int solve(vector<int>& nums, int prev, int n, int mask, unordered_set<int> &squares, map<pair<int,int>, int> &dp){
        if(__builtin_popcount(mask) == n) return 1;

        // if(dp.count({mask, prev})) return dp[{mask, prev}];

        int ans = 0;
        unordered_map<int, bool> mp;

        for(int i = 0; i < n; i++){
            if(mask & (1<<i)) continue;
            if(squares.count(prev + nums[i]) && !mp.count(nums[i])){
                ans += solve(nums, nums[i], n, (mask | (1<<i)), squares, dp);
                mp[nums[i]] = 1;
            }
        }

        // return dp[{mask, prev}] = ans;
        return ans;
    }

public:
    int numSquarefulPerms(vector<int>& nums) {
        int n = nums.size(), ans = 0;
        int mx = *max_element(nums.begin(), nums.end());
        unordered_set<int> squares;
        map<pair<int,int>, int> dp;

        for(int i = 0; i*i <= 2*mx; i++){
            squares.insert(i*i);
        }
        
        unordered_map<int, bool> mp;
        for(int i = 0; i < n; i++){
            if(!mp.count(nums[i])){
                ans += solve(nums, nums[i], n, (1<<i), squares, dp);
                mp[nums[i]] = 1;
            } 
        }

        return ans;
    }
};
46500