<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)
    #     temp = []
    #     result = set()

    #     def dfs(index):
    #         if index == n:
    #             if len(temp) >= 2:
    #                 result.add(tuple(temp))
    #             return
            
    #         if not temp or nums[index] >= temp[-1]:
    #             temp.append(nums[index])
    #             dfs(index+1)
    #             temp.pop(-1)
    #         dfs(index+1)

    #     dfs(0)
    #     return [*result]

    def findSubsequences(self, nums: List[int]) -> List[List[int]]:
        n = len(nums)

        def solve(index):
            if index == n:
                return [()]
            
            r = solve(index + 1)
            _r = []
            for s in r:
                if not s:
                    _r.append((index,))
                elif nums[index] <= nums[s[0]]:
                    _r.append((index, *s))
            _r += r
            return _r

        r = solve(0)
        result = set()
        for s in r:
            l = [ nums[i] for i in s ]
            if len(l) >= 2:
                result.add(tuple(l))
        
        return result

        
            
            


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