<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]]:
        path = []
        result = []

        len_nums = len(nums)

        def backtracking(start_idx):
            if len(path) >= 2:
                result.append(path.copy())

            if start_idx >= len_nums:
                return 
            
            used = set()
            for idx in range(start_idx, len_nums):
                if path and nums[idx] < path[-1]:
                    continue 
                if nums[idx] in used:
                    continue
                
                used.add(nums[idx])
                path.append(nums[idx])
                backtracking(idx + 1)
                path.pop()

        backtracking(0)
        return result
                
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