<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>> ans;
    set<vector<int>> st;
    void backtrack(vector<int>& nums, int i, vector<int> temp) {
        if (temp.size() >= 2) {
            if (st.find(temp) == st.end()) {
                ans.push_back(temp);
                st.insert(temp);
            }
        }
        if (i == nums.size()) {
            return;
        }
        for (int j = i; j < nums.size(); j++) {
            if (!temp.empty() && nums[j] < temp.back()) {
                continue;
            }
            temp.push_back(nums[j]);
            backtrack(nums, j + 1, temp);
            temp.pop_back();
        }
    }
    vector<vector<int>> findSubsequences(vector<int>& nums) {
        ios::sync_with_stdio(false);
        cin.tie(nullptr);
        cout.tie(nullptr);
        backtrack(nums, 0, {});
        return ans;
    }
};
37671