<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 solve(vector<int> &nums, set<vector<int>> &s, int idx, vector<int> temp){
        if(temp.size()>=2){
            s.insert(temp);
        }
        
        if(idx>= nums.size()){
            return;
        }

        if(temp.empty() || nums[idx]>= temp.back()){
            temp.push_back(nums[idx]);
            solve(nums, s, idx+1, temp); // take the element
            temp.pop_back();
        }
        solve(nums, s, idx+ 1, temp);  //do not take the element
    }

    vector<vector<int>> findSubsequences(vector<int>& nums) {
        set<vector<int>> s;
        vector<int> temp; 
        solve(nums, s, 0, temp);
        vector<vector<int>> ans;

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