<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:
    set<vector<int>> ans;

    void helper(vector<int> temp_ans,int i,vector<int>& nums)
    {
       
        if(temp_ans.size() >= 2)
        {
            ans.insert(temp_ans);

        }
         if(i == nums.size())
            return;

         helper(temp_ans,i+1,nums);
       
        if(temp_ans.size()==0 || temp_ans[temp_ans.size() - 1] <= nums[i])
        {
            temp_ans.push_back(nums[i]);
            helper(temp_ans,i+1,nums);

        }
    
    
          


        return;
    }

    vector<vector<int>> findSubsequences(vector<int>& nums) {

        vector<int> temp_ans;
        helper(temp_ans,0,nums);
        vector<vector<int>> orig_ans;

        for(auto s : ans)
        {
            orig_ans.push_back(s);
        }

        return orig_ans;
        
    }
};
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