<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:
    def findSubsequences(self, nums: List[int]) -> List[List[int]]:
        """
        Steps:
        - Iterate over each element
            - Consider to include or exclude each element
            - Only include when the current value is greater than
                previous value
            - Only add to result if the size is 2 or greater
        """

        result = set()
        slate = []

        def dfs(idx: int, prev: int) -> None:

            # leaf 
            if idx == len(nums):
                if len(slate) >= 2:
                    result.add(tuple(slate))
                return

            # inner
            # exclude
            dfs(idx+1, prev)

            # include
            if nums[idx] >= prev:
                slate.append(nums[idx])
                dfs(idx+1, nums[idx])
                slate.pop()
        
        dfs(idx=0, prev=float('-inf'))

        return list(result)

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