<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) {
        map<int,vector<vector<int>>>m;
        map<pair<int,int>,int>v;
        for(int i=1;i<nums.size();i++)
        {
            int j=i-1;
            while(j>=0)
            {
                if(nums[i]>=nums[j])
                {
                    if(v.find({nums[j],nums[i]})==v.end())
                    m[i].push_back({nums[j],nums[i]});
                    v[{nums[j],nums[i]}]++;
                    for(auto it:m[j])
                    {
                        vector<int>y=it;
                        y.push_back(nums[i]);
                        m[i].push_back(y);
                    }
                }
                if(nums[i]==nums[j])
                break;
                    j--;
            }
        }
        vector<vector<int>>ans;
        
        for(auto it:m)
        {
            for(auto x:it.second)
            {
                ans.push_back(x);
            }
        }
        return ans;
    }
};
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