<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]]:
        
        def helper(i):
            if i == len(nums):
                if len(slate) >= 2:
                    result.add(tuple(slate[:]))
                return
                
            counter = 1
            j = i
            while (j + 1) < len(nums) and nums[j] == nums[j + 1]:
                counter += 1
                j += 1

            # Option 1: Skip all duplicates of nums[i]
            helper(i + counter)
            
            # Option 2: Include duplicates of nums[i]
            if len(slate) == 0 or nums[i] >= slate[-1]:
                for k in range(1, counter + 1):
                    # Include k copies of nums[i]
                    slate.extend([nums[i]] * k)
                    helper(i + counter)
                    # Backtrack by removing the added nums[i]
                    for _ in range(k):
                        slate.pop()
        
        result = set()
        slate = []
        helper(0)
        return [list(seq) for seq in result]

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