<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]]:
        """
        - 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) -> None:
            #leaf
            if idx == len(nums):
                if len(slate) >= 2:
                    result.add(tuple(slate))
                return

            # exclude
            dfs(idx+1)

            #include
            if len(slate) == 0 or nums[idx] >= slate[-1]:
                slate.append(nums[idx])
                dfs(idx+1)
                slate.pop()

        dfs(idx=0)

        return list(result)
25200