<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)
        res = []
        path = []

        def dfs(idx): # idx: to locate children
            # when do we collect result:
            # exhaust all paths but only collect when at least 2 numbers

            # don't include this node
            
            if len(path) >= 2:
                res.append([nums[i] for i in path])

            # BUG: handle duplicate
            used = set()
            
            for i in range(idx, n):
                if nums[i] in used:
                    continue
                used.add(nums[i])
                
                # if (i > idx and nums[i] <= nums[i-1]):
                #     continue
                
                # if len(path) > 0 and nums[i] < path[-1]: # BUG: path stores index!!
                if len(path) > 0 and nums[i] < nums[path[-1]]: # handle non-decreasing
                    continue

                path.append(i) # include this node
                dfs(i+1)
                path.pop()
        
        dfs(0)

        return res
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