<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 Node{
public:
    unordered_map<int, Node*> mp;
    bool vis;
    Node(){
        vis = false;
    }
    bool add(int num){        
        if(mp.find(num) == mp.end()){
            mp[num] = new Node();
            return true;
        }
        return false;
    }
};

class Solution {
public:
    vector<vector<int>> res;
    void backtrack(vector<int>& nums, vector<int>& cur, int idx, Node* node){
        if(idx >= nums.size()){
            if(cur.size() >= 2 && !node->vis){
                res.push_back(cur);
                node->vis = true;
            }
            return;
        }
        Node* prev = node;
        if(cur.empty() || nums[idx] >= cur[cur.size()-1]){
            cur.push_back(nums[idx]);
            node->add(nums[idx]);
            node = node->mp[nums[idx]];
            backtrack(nums, cur, idx+1, node);
            cur.pop_back();
            node = prev;
        }
        backtrack(nums, cur, idx+1, node);
    }
    vector<vector<int>> findSubsequences(vector<int>& nums) {
        vector<int> cur;
        Node* node = new Node();
        backtrack(nums, cur, 0, node);
        return res;
    }
};
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