<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>> result;
    vector<int> subset;
    void backtracking(vector<int>& nums, int numsindex){
        if(subset.size()>1) result.push_back(subset);
        for(int i=numsindex; i<nums.size(); i++){
            if(i>numsindex && find(nums.begin()+numsindex, nums.begin()+i, nums[i])!=nums.begin()+i) continue;
            if(subset.size()==0 || (subset.size()>0 && nums[i]>=*(subset.end()-1))){
                subset.push_back(nums[i]);
                backtracking(nums, i+1);
                subset.pop_back();
            }
        }
    }
    vector<vector<int>> findSubsequences(vector<int>& nums) {
        backtracking(nums, 0);
        return result;
    }
};
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