<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:
    void dfs(const int n, unordered_map<int, int>& remaining, const unordered_set<int>& squares, long& output, int idx, int last){
        if (idx == n){
            output += 1;
            return;
        }
        for (auto& [k, v] : remaining){
            if (v == 0) continue;
            if (idx == 0 || (idx != 0 && squares.contains(k + last))) {
                remaining[k]--;
                dfs(n, remaining, squares, output, idx+1, k);
                remaining[k]++;
            }
        }
    }
    int numSquarefulPerms(vector<int>& nums) {
        const int n = nums.size();
        vector<int> seen(n, 0);
        unordered_set<int> squares;
        long max = *max_element(nums.begin(), nums.end());
        for (int i = 0; i < 1e6; i++){
            long square = i*i;
            if (square < 2*max+1) {
                squares.insert(square);
            } else {
                break;
            }
        }
        long out = 0;
        unordered_map<int, int> counts;
        for (int i = 0; i < n; i++){
            counts[nums[i]]++;
        }
        dfs(n, counts, squares, out, 0, 0);
        return out;
    }
};
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