<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:
    void solve(vector<int> &nums, vector<int> &temp, vector<vector<int>> &ans, int idx) {
        if (temp.size() > 1) {
            ans.push_back(temp);
        }

        set<int> used;  // Local set to avoid duplicates at this level

        for (int i = idx; i < nums.size(); ++i) {
            if ((temp.empty() || temp.back() <= nums[i]) && used.find(nums[i]) == used.end()) {
                used.insert(nums[i]);  // Mark this element as used in this level
                temp.push_back(nums[i]);
                solve(nums, temp, ans, i + 1);
                temp.pop_back();  // Backtrack
            }
        }
    }

    vector<vector<int>> findSubsequences(vector<int>& nums) {
        vector<int> temp;
        vector<vector<int>> ans;
        solve(nums, temp, ans, 0);
        return ans;
    }
};

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