<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 n;

    // Check if a number is a perfect square
    bool checkPerfect(long long num) {
        long long root = sqrt(num);
        return (root * root == num);
    }

    // Helper function to count squareful permutations
    int helper(int count, int mask, int prevInd, vector<int>& nums, vector<vector<int>>& dp) {
        // Base case: if all numbers are used
        if (mask == (1 << n) - 1) {
            return 1;
        }
        // Use memoization to return already computed results
        if (dp[mask][prevInd + 1] != -1) {
            return dp[mask][prevInd + 1];
        }
        int ans = 0;
        for (int i = 0; i < n; i++) {
            // Skip already used elements
            if (mask & (1 << i)) {
                continue;
            }
            // Skip duplicates
            if (i > 0 && nums[i] == nums[i - 1] && !(mask & (1 << (i - 1)))) {
                continue;
            }
            // Recursively count valid permutations
            if (prevInd == -1 || checkPerfect((long long)nums[i] + nums[prevInd])) {
                ans += helper(count + 1, mask | (1 << i), i, nums, dp);
            }
        }
        return dp[mask][prevInd + 1] = ans;
    }

public:
    int numSquarefulPerms(vector<int>& nums) {
        n = nums.size();
        sort(nums.begin(), nums.end()); // Sort the numbers to handle duplicates
        vector<vector<int>> dp((1 << n), vector<int>(n + 1, -1)); // DP table for memoization
        return helper(0, 0, -1, nums, dp);
    }
};
16800