<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:
    vector<vector<int>> findSubsequences(vector<int>& nums) {
        vector<int> res;
        vector<vector<int>> ans;
        set<vector<int>> s;
        int n=nums.size();
        fill(0, n, nums, res, ans,s);
        return ans;
    }

    void fill(int i, int n, vector<int> nums, vector<int> &res, vector<vector<int>> &ans, set<vector<int>> &s){
        if(res.size()>1){
            if(s.find(res)==s.end()){
                    ans.push_back(res);
            s.insert(res);
            }
            
        }
        if(i==n) return;
        
        if(res.empty() || (!res.empty() && nums[i]>=res.back())){
            res.push_back(nums[i]);
            fill(i+1, n, nums, res, ans,s);
            res.pop_back();
        }
        fill(i+1, n, nums, res, ans,s);
    }
};
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