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
func findSubsequences(nums []int) [][]int {
	var ans [][]int
	var dfs func(u, last int, t []int)
	dfs = func(u, last int, t []int) {
		if u == len(nums) {
			if len(t) > 1 {
				ans = append(ans, slices.Clone(t))
			}
			return
		}
		if nums[u] >= last {
			t = append(t, nums[u])
			dfs(u+1, nums[u], t)
			t = t[:len(t)-1]
		}
		if nums[u] != last {
			dfs(u+1, last, t)
		}
	}
	var t []int
	dfs(0, -1000, t)
	return ans
}