<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>> ans;
    set<vector<int>> s;
    void find(vector<int> &nums,vector<int> arr,int n,int ind)
    {
        if(ind == n)
        {
            if(arr.size()>=2 && s.find(arr) == s.end())
            {
                 ans.push_back(arr);
                 s.insert(arr);
            }
            return;
        }

        if(arr.size() == 0)
        {
            find(nums,arr,n,ind+1);
            arr.push_back(nums[ind]);
            find(nums,arr,n,ind+1);
            arr.pop_back();
        }
        else{
            find(nums,arr,n,ind+1);
            if(arr[arr.size()-1]<=nums[ind])
            {
                arr.push_back(nums[ind]);
                find(nums,arr,n,ind+1);
                arr.pop_back();
            }

        }

        return;
    }

    vector<vector<int>> findSubsequences(vector<int>& nums) {
        vector<int> arr;
        int n = nums.size();
        find(nums,arr,n,0);
        return ans;  
    }
};
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