<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]]:
        # Use a set to store unique subsequences
        output = set()
        
        def dfs(i, sequence):
            # If we've reached the end of the array
            if i == len(nums):
                if len(sequence) > 1:
                    output.add(tuple(sequence))  # Store as tuple to avoid duplicates
                return 

            # Option 1: Include the current element if valid (non-decreasing condition)
            if not sequence or nums[i] >= sequence[-1]:
                sequence.append(nums[i])  # Choose the current element
                dfs(i + 1, sequence)  # Explore with the current element
                sequence.pop()  # Backtrack by removing the element
                
            # Option 2: Don't include the current element
            dfs(i + 1, sequence)

        # Start DFS from the first index
        dfs(0, [])
        
        # Convert the set back to a list of lists for the final output
        return list(output)

25200