<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<int> curr;
        vector<vector<int>> res;
        backtrack(0, curr, nums, res);
        return res;
    }
private:
    void backtrack(int start, vector<int>& curr, vector<int>& nums, vector<vector<int>>& res){
        if (curr.size() >= 2)
            res.push_back(curr);
        if (start == nums.size())
            return;
        bool seen[201] = {};
        for (int i = start; i < nums.size(); i++){
            if (seen[nums[i] + 100])
                continue;
            else if (curr.empty() || nums[i] >= curr.back()){
                curr.push_back(nums[i]);
                backtrack(i + 1, curr, nums, res);
                curr.pop_back();
                seen[nums[i] + 100] = true;
            }
        }
    }
};
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