<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 cnt = 0 ;
    vector<int> v;
    vector<vector<int>> res;
    void dfs(int idx,vector<int>& nums){
        if(idx == nums.size()){
            return;
        }
        for(int i = idx; i < nums.size(); i++){
            v.push_back(nums[i]);
            res.push_back(v); 
            dfs(i+1,nums);
            v.pop_back();
        }
    }
    int beautifulSubsets(vector<int>& nums, int k) {
        dfs(0,nums);
        int n = res.size();
        for(int i = 0 ; i < n ; i++){
            bool check = true;
            for(int j = 0 ; j < res[i].size();j++){
                for(int l = j+1 ; l < res[i].size(); l++){
                    int x = abs(res[i][l] - res[i][j]);
                    if(x == k){
                        check = false;
                        break;
                    }
                }
                if(!check){
                    break;
                }
            }
            if(check) cnt++;
        }
        return cnt;
    }
};

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