<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;
    vector<int> sub;
    void solve(vector<int> nums, int i){
        if(sub.size()>1){
            ans.insert(sub);
        }
        if(i>=nums.size()) return;
        if(!sub.empty() && sub.back()<=nums[i]){
            sub.push_back(nums[i]);
            solve(nums, i+1);
            sub.pop_back();
        }
        solve(nums, i+1);
            
    }
    vector<vector<int>> findSubsequences(vector<int>& nums) {
        for(int i=0; i<nums.size(); i++){
            sub.push_back(nums[i]);
            solve(nums, i+1);
            sub.pop_back();
        }
        vector<vector<int>> v;
        for(auto i: ans) v.push_back(i);
        return v;
    }
};
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