<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>>  map;
    vector<vector<int>> answer;
    vector<int> sub;
    void sum(vector<int>& arr,int index,int n){
        if(index>=n){
            return;
        }
        
        if(sub.size()==0 || arr[index]>=sub.back()){
            sub.push_back(arr[index]);
            if(sub.size()>=2 && map.find(sub)==map.end()){
                map.insert(sub);
                answer.push_back(sub);
            }

            sum(arr,index+1,n);
            sub.pop_back();
        }
        sum(arr,index+1,n);

    }
    vector<vector<int>> findSubsequences(vector<int>& nums) {
        // sort(nums.begin(),nums.end());
        sum(nums,0,nums.size());
        return answer;
    }

};
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