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:
    def findSubsequences(self, nums: List[int]) -> List[List[int]]:
        def dfs(u, last, t):
            if u == len(nums):
                if len(t) > 1:
                    ans.append(t[:])
                return
            if nums[u] >= last:
                t.append(nums[u])
                dfs(u + 1, nums[u], t)
                t.pop()
            if nums[u] != last:
                dfs(u + 1, last, t)

        ans = []
        dfs(0, -1000, [])
        return ans
