<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 f(int ind,vector<int>& nums,set<vector<int>>&st,vector<int>&vec)
   {
    int n=nums.size();
    if(ind>=n){
        if(vec.size()>1)
        {
            if(st.find(vec)==st.end())
             st.insert(vec);
        }
        return;
    }

    if(vec.empty())
    {
        f(ind+1,nums,st,vec);
        vec.push_back(nums[ind]);
        f(ind+1,nums,st,vec);
        vec.pop_back();
    }
    else 
    {
        if(nums[ind]<vec.back())
            f(ind+1,nums,st,vec);
        else 
        {
          f(ind+1,nums,st,vec);
          vec.push_back(nums[ind]);
          f(ind+1,nums,st,vec);
         vec.pop_back();  
        }
    }
   }
    vector<vector<int>> findSubsequences(vector<int>& nums) {
        vector<vector<int>>res;
        int n=nums.size();
        vector<int>vec;
        set<vector<int>>st;
        f(0,nums,st,vec);
        for(auto it:st )
        res.push_back(it);
        return res;
    }
};
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