<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 buildSequence(self,startIndex, path):
        
        if len(path) >= 2:
            self.res.add(tuple(path))

        for j in range(startIndex + 1, len(self.nums)):

            if self.nums[j] < path[-1]:
                continue
            path.append(self.nums[j])
            self.buildSequence(j, path)
            path.pop()

    def findSubsequences(self, nums: List[int]) -> List[List[int]]:
        if len(nums) < 2:
            return []

        self.nums = nums
        self.res = set()
        self.used = [False] * len(self.nums)

        for i in range(len(nums) - 1):
            self.buildSequence(i, [nums[i]])
        
        return list(self.res)

23000