<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 solve(int index,vector<int> &nums,int k,vector<int> curr)
    {
        if(index==nums.size())
        {
            if(curr.size()>0)
                return 1;
            else
                return 0;
        }
        int ans=0;
        if(curr.size()>0)
            ans+=1;
        for(int i=index;i<nums.size();i++)
        {
            if(curr.size()==0)
            {
                curr.push_back(nums[i]);
                ans+=solve(i+1,nums,k,curr);
                curr.pop_back();
            }
            else
            {
                int flag=0;
                for(auto j:curr)
                {
                    if(abs(j-nums[i])==k)
                    {
                        flag=1;
                        break;
                    }
                }
                if(flag==0)
                {
                    curr.push_back(nums[i]);
                    ans+=solve(i+1,nums,k,curr);
                    curr.pop_back();
                }
            }
        }
        return ans;
            
    }
    int beautifulSubsets(vector<int>& nums, int k) {
        return solve(0,nums,k,{});
    }
};
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