<p>You are given an array <code>nums</code> of positive integers and a <strong>positive</strong> integer <code>k</code>.</p>

<p>A subset of <code>nums</code> is <strong>beautiful</strong> if it does not contain two integers with an absolute difference equal to <code>k</code>.</p>

<p>Return <em>the number of <strong>non-empty beautiful </strong>subsets of the array</em> <code>nums</code>.</p>

<p>A <strong>subset</strong> of <code>nums</code> is an array that can be obtained by deleting some (possibly none) elements from <code>nums</code>. Two subsets are different if and only if the chosen indices to delete are different.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [2,4,6], k = 2
<strong>Output:</strong> 4
<strong>Explanation:</strong> The beautiful subsets of the array nums are: [2], [4], [6], [2, 6].
It can be proved that there are only 4 beautiful subsets in the array [2,4,6].
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [1], k = 1
<strong>Output:</strong> 1
<strong>Explanation:</strong> The beautiful subset of the array nums is [1].
It can be proved that there is only 1 beautiful subset in the array [1].
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 20</code></li>
	<li><code>1 &lt;= nums[i], k &lt;= 1000</code></li>
</ul>

class Solution {
public:
    int beautifulSubsets(vector<int>& nums, int k) {
        int n = nums.size();
        vector<vector<int>> dp(n + 1, vector<int>(1 << n, -1)); // dp table to store results
        sort(nums.begin(), nums.end()); // Sort nums to ensure we handle subsets properly

        return countBeautifulSubsets(nums, k, 0, 0, dp) - 1; // Subtract 1 to exclude the empty subset
    }

private:
    int countBeautifulSubsets(vector<int>& nums, int k, int index, int mask, vector<vector<int>>& dp) {
        if (index == nums.size()) {
            return 1; // Return 1 for each valid subset (including empty subset initially)
        }

        if (dp[index][mask] != -1) {
            return dp[index][mask];
        }

        // Exclude current element
        int result = countBeautifulSubsets(nums, k, index + 1, mask, dp);

        // Include current element if it does not violate the beautiful subset condition
        bool canInclude = true;
        for (int i = 0; i < nums.size(); ++i) {
            if (mask & (1 << i)) { // Check if element i is included in the current subset
                if (abs(nums[index] - nums[i]) == k) {
                    canInclude = false;
                    break;
                }
            }
        }

        if (canInclude) {
            result += countBeautifulSubsets(nums, k, index + 1, mask | (1 << index), dp);
        }

        return dp[index][mask] = result;
    }
};

209822