<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]]:
        n = len(nums)  # Properly indenting this
        result = []    # Result list inside the method
        
        # Backtracking function
        def backtrack(start, current):
            used = set()  # Set to keep track of used numbers at each level
            if len(current) >= 2:
                result.append(current[:])  # Add a copy of current subsequence if valid
            for i in range(start, n):
                if nums[i] in used:  # Skip if the number has been used at this level
                    continue
                if not current or nums[i] >= current[-1]:  # Ensure increasing order
                    current.append(nums[i])  # Choose the current number
                    backtrack(i + 1, current)  # Explore further
                    current.pop()  # Undo the choice (backtrack)
                    used.add(nums[i])  # Mark this number as used at this level
        
        # Start backtracking
        backtrack(0, [])
        return result

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