Non-decreasing Subsequences
<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>
Bit Manipulation; Array; Hash Table; Backtracking
class Solution {
public:
    vector<vector<int>> findSubsequences(vector<int>& nums) {
        vector<vector<int>> ans;
        vector<int> t;
        dfs(0, -1000, t, nums, ans);
        return ans;
    }

    void dfs(int u, int last, vector<int>& t, vector<int>& nums, vector<vector<int>>& ans) {
        if (u == nums.size()) {
            if (t.size() > 1) ans.push_back(t);
            return;
        }
        if (nums[u] >= last) {
            t.push_back(nums[u]);
            dfs(u + 1, nums[u], t, nums, ans);
            t.pop_back();
        }
        if (nums[u] != last) dfs(u + 1, last, t, nums, ans);
    }
};