<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:
    set<vector<int>> ans;
    void recur(vector<int>& nums, int ind, int n, vector<int> v) {
        if(ind >= n) return;

        for(int i = ind; i < n; i++) {
            // if(i > 0 && nums[i] == nums[i-1]) continue;
            int p = v.size();
            if(p > 0 && v[p-1] > nums[i]) continue;

            v.push_back(nums[i]);
            recur(nums, i + 1, n, v);
            if(v.size() > 1) ans.insert(v);
            v.pop_back();
        }
    }

    vector<vector<int>> findSubsequences(vector<int>& nums) {
        ans.clear();
        vector<int> curr;
        vector<vector<int>> subSeqs;
        recur(nums, 0, nums.size(), curr);

        auto it = ans.begin();
        while(it != ans.end()) {
            subSeqs.push_back(*it);
            it++;
        }
        return subSeqs;
    }
};
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