<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 help(map<int,int>& arr,int k){
    int next1=1;int next2=1;
    for(auto it=arr.rbegin();it!=arr.rend();it++){
        int pick=0;
        
        if(it!=arr.rbegin()){
            if(arr.find(it->first+k)!=arr.end()){
               pick=(pow(2,it->second)-1)*next2;
            }
            else{
              pick=(pow(2,it->second)-1)*next1;
            }
        }
        else{
            pick=(pow(2,it->second)-1);
        }
        int notpick=next1;
        int curr=pick+notpick;
        next2=next1;
        next1=curr;
    }
    return next1;
   }
    int beautifulSubsets(vector<int>& nums, int k) {
        int ans=1;
        sort(nums.begin(),nums.end());
        unordered_map<int,map<int,int>> mp;
        for(int i=0;i<nums.size();i++){
            mp[nums[i]%k][nums[i]]++;
        }
        for(int i=0;i<k;i++){
            int res=help(mp[i],k);
            // cout<<i<<" "<<res<<endl;
            ans=ans*res;
        }
        return ans-1;
    }
};
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