<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>> res;
        set<vector<int>> s;
        int n=nums.size();
        for(int i=0;i<(1<<n);i++){
            vector<int> temp;
            int prev=INT_MIN,flag=0;
            for(int j=0;j<n;j++){
                if(i & (1<<j)){
                    if(nums[j]<prev){
                        flag=1;
                        break;
                    }
                    prev=nums[j];
                    temp.push_back(nums[j]);
                }
            }
            if(flag==0 && temp.size()>=2){
                //res.push_back(temp);
                s.insert(temp);
            }
        }
        for(auto each: s){
            res.push_back(each);
        }
        //sort(res.begin(),res.end());
        return res;
    }
};
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