<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:
    unordered_map<string, bool> map;
    void helper (vector<int>& arr, int start, vector<vector<int>> &ans, vector<int> &curr) {
        if (start == arr.size() && curr.size() >= 2) {
            string s = "";
            for(int i=0; i<curr.size(); i++) {
                s += to_string(curr[i]) + ",";
            }
            if (!map[s]) {
                ans.push_back(curr);
                map[s] = true;
            }
            return;
        }

        if (start >= arr.size()) {
            return;
        }

        helper(arr, start + 1, ans, curr);
        if (curr.size() == 0 || curr[curr.size() -1] <= arr[start]) {
            curr.push_back(arr[start]);
            helper(arr, start + 1, ans, curr);
            curr.pop_back();
        }
    }

    vector<vector<int>> findSubsequences(vector<int>& arr) {
       vector<vector<int>> ans;
       vector<int> curr;
       helper(arr, 0, ans, curr);
       return ans;
    }
};
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