<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) {
        int size = nums.size();
        vector<vector<int>> table;
        vector<vector<int>> ans;
        for(int i=0; i<size; i+=1){
            int len = table.size();
            for(int j=len-1; j>=0; j-=1){
                if(table[j].back()<=nums[i]){
                    table.push_back(table[j]);
                    table.back().push_back(nums[i]);
                }
            }
            table.emplace_back(1, nums[i]);
        }

        vector<int> order(table.size());
        for(int i=0; i<table.size(); i+=1){
            order[i] = i;
        }
        sort(order.begin(), order.end(), [&table](int a, int b){
            return table[a]<table[b];
        });
        for(int i=0; i<order.size(); i+=1){
            if(table[order[i]].size()==1){
                continue;
            }
            if(i>0 && table[order[i]]==table[order[i-1]]){
                continue;
            }
            ans.push_back(table[order[i]]);
        }

        return ans;
    }
};
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