<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:

    /**s contiene gli indici degli elementi appartenenti alla subset
    subset distinte !!!**/
    void dfs(vector<int> &s,int x,vector<int> &A,int k,int &number,vector<int> &cnt){
        
        vector<int> candidates;

        /**se la differenza tra cnt[A[i] -k] è maggiore di 0 vuol dire 
        che non lo posso inserire,altrimenti se vale 0 lo posso includere**/
        int pos = (x==0) ? 0 : s[x-1]+1;
            
        for(int i=pos;i<A.size();i++){
            if((A[i] - k) < 0){
                if(cnt[A[i]+k] == 0){
                    candidates.push_back(i);
                }
            }else{
                if(cnt[A[i]-k]==0 && cnt[A[i]+k] == 0){
                    candidates.push_back(i);
                }
            }
        }

        x++;
        for(int i=0;i<candidates.size();i++){
            s.push_back(candidates[i]);
            cnt[A[s[x-1]]]++;
            number++;
            dfs(s,x,A,k,number,cnt);
            cnt[A[s[x-1]]]--;
            s.pop_back();
        }
        return ;
    }


    int beautifulSubsets(vector<int>& nums, int k) {
      
        sort(nums.begin(),nums.end());
        vector<int> s;
        int number = 0;

        vector<int> cnt(2001,0);

        dfs(s,0,nums,k,number,cnt);

        return number;
    }
};
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