<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:
    bool valid(vector<int> &arr, int k)
    {
        for (int i = 0; i < arr.size(); i++)
        {
            for (int j = i + 1; j < arr.size(); j++)
            {
                if (abs(arr[i] - arr[j]) == k) return false;
            }
        }
        return true;
    }
    void backtrack(vector<int> &nums, vector<vector<int>> &sum, vector<int> &arr, int k, vector<int> &visited, int t)
    {
        if (arr.size() == nums.size()) return;

        for (int i = t; i < nums.size(); i++)
        {
            if (visited[i]) continue;
            visited[i] = 1;
            arr.push_back(nums[i]);
            if (valid(arr, k))
            {
                sum.push_back(arr);
                backtrack(nums, sum, arr, k, visited, i);
            }
            arr.pop_back();
            visited[i] = 0;
        }
    }
    int beautifulSubsets(vector<int>& nums, int k) {
        vector<vector<int>> sum;
        vector<int> arr;
        vector<int> visited(nums.size(), 0);
        backtrack(nums, sum, arr, k, visited, 0);
        // for (vector<int> &a : sum)
        // {
        //     for (int i : a) cout << i << " ";
        //     cout << endl;
        // }
        return sum.size();
    }
};
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