<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(int ind,vector<int> nums,vector<int> &temp,set<vector<int>> &ans){
        if(temp.size()>=2){
            ans.insert(temp);
        }
        if(ind>=nums.size())    
                return;
        for(int i=ind;i<nums.size();i++){
           if(temp.empty() || nums[i]>=temp.back())
            {
                temp.push_back(nums[i]);
                solve(i+1,nums,temp,ans);
                temp.pop_back();
            }
        }
    }
    vector<vector<int>> findSubsequences(vector<int>& nums) {
        set<vector<int>> ans;
        vector<vector<int>> res;
        vector<int> temp;
        solve(0,nums,temp,ans);
        for(vector<int> t: ans){
            res.push_back(t);
        }
        return res;
    }
};
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