<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) {
        set<vector<int>> res;
        int sz = nums.size();
        vector<vector<vector<int>>> dp(sz);
        for(int i = 0; i < sz; i++) {
            dp[i].push_back({nums[i]});
            for(int j = i - 1; j >= 0; j--) {
                if(nums[i] >= nums[j]) {
                    for(auto c : dp[j]) {
                        c.push_back(nums[i]);
                        res.insert(c);
                        dp[i].push_back(c);
                    }
                }
            }
        }
        vector<vector<int>> rep;
        for(auto itr : res)
            rep.push_back(itr);
        return rep;
    }
};
46801