<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>> result;
        vector<int> temp;
        dfs(nums, result, temp, 0);
        return result;
    }

private:
    void dfs(vector<int>& nums, vector<vector<int>>& result, vector<int>& temp, int index)
    {
        if (temp.size() > 1)
        {
            if (find(result.begin(), result.end(), temp) == result.end())
            result.push_back(temp);
        }

        for (int i = index; i < nums.size(); ++i)
        {
            int back = temp.empty() ? -101 : temp.back();

            if (nums[i] >= back)
            {
            temp.push_back(nums[i]);
            dfs(nums, result, temp, i + 1);
            temp.pop_back();
            }
        }      
    }
};
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