<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 {
private:
    void getSequences(int index,vector<int>& inter,vector<vector<int>>& ans,vector<int>& nums){
        if(inter.size()>=2 && index!=nums.size()){
            ans.push_back(inter);
        }
        if(index==nums.size()){
            if(inter.size()>=2){
                ans.push_back(inter);
            }
            return ;
        }
        map<int,int>mpp;
        for(int i=index;i<nums.size();i++){
            
            if(mpp.find(nums[i])!=mpp.end())continue;

            if((inter.size()==0 || nums[i] >= inter.back() ) ){
                inter.push_back(nums[i]);
                mpp[nums[i]]=1;
                getSequences(i+1,inter,ans,nums);
                inter.pop_back();
            }
        }
    }
public:
    vector<vector<int>> findSubsequences(vector<int>& nums) {
        vector<vector<int>>ans;
        vector<int>inter;
        int n=nums.size();
        getSequences(0,inter,ans,nums);
        return ans;
    }
};
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