<p>Given an integer array <code>nums</code>, return <em>all the different possible non-decreasing subsequences of the given array with at least two elements</em>. You may return the answer in <strong>any order</strong>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [4,6,7,7]
<strong>Output:</strong> [[4,6],[4,6,7],[4,6,7,7],[4,7],[4,7,7],[6,7],[6,7,7],[7,7]]
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [4,4,3,2,1]
<strong>Output:</strong> [[4,4]]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 15</code></li>
	<li><code>-100 &lt;= nums[i] &lt;= 100</code></li>
</ul>

class Solution {
public:
  void rec(vector<int>& nums,int i, vector<int> v, set<vector<int>>& s){
           s.insert(v);

           for(int j=i;j<nums.size();j++){
           if(v.empty() || nums[j]>=v.back()){
           v.push_back(nums[j]);}
           rec(nums,j+1,v,s);
           if(!v.empty()){
            v.pop_back();
           } 
           }

    //        for(int j=i;j<nums.size();j++){

    //     if(j!=i && nums[j]==nums[j-1]){continue;}
    //     v.push_back(nums[j]);
    //     rec(nums,);
    //     v.pop_back();
    // }

  }
    vector<vector<int>> findSubsequences(vector<int>& nums) {
        set<vector<int>> s;
        vector<int> v;
        rec(nums,0,v,s);

        vector<vector<int>> ans;
        for(auto& it: s){
            if(it.size()>=2){
            ans.push_back(it);}
            }

        return ans;
    }
};
50225