<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<vector<int>> answer;
        set<vector<int>> s;
        int x = pow(2,nums.size());
        for (int i=0;i<x;i++){
            int y=i;
            int j=0;
            if ((i&(i-1))==0 || i==0) continue;
            vector<int> count;
            while(y>0){
                if (y%2==1) {
                    if (count.empty()) count.push_back(nums[j]);
                    else{
                        if (count.back()<=nums[j]) count.push_back(nums[j]);
                    }
                }
                j++;
                y/=2;
            }
            if (count.size()>=2) s.insert(count);
        }
        for (auto x:s){
            answer.push_back(x);
        }
        return answer;
    }
};
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