<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:
    int n;
    vector<vector<int>> findSubsequences(vector<int>& nums) {
        n = nums.size();
        if(n == 1) return {};
        vector<vector<int>> result;
        vector<int> path;
        backtrack(result, path, nums, 0);
        return result;

    }

    void backtrack(vector<vector<int>>& result, vector<int>& path, vector<int>& nums, int idx) {
        if(path.size() >= 2) {
            result.push_back(path);
        }
        map<int, bool> mp;
        for(int i = idx; i<n; ++i) {
            if(mp[nums[i]]) continue;
            else if(path.empty()  || nums[i] >= path.back()) {
                path.push_back(nums[i]);
                mp[nums[i]] = true;
                backtrack(result, path, nums, i+1);
                path.pop_back();
            }
        }
    }
};
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