<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]]:
        if len(nums) <= 1:
            return []
        
        outputs = {}
        for i in range(len(nums)):
            outputs_begin_idx_i = [[nums[i]]]
            for j in range(i + 1, len(nums)):
                new_arrs = []
                for arr in outputs_begin_idx_i:
                    if nums[j] >= arr[-1]:
                        new_arrs.append(
                            arr + [nums[j]]
                        )
                outputs_begin_idx_i.extend(new_arrs)
            for arr in outputs_begin_idx_i:
                outputs[str(arr)] = arr
        
        output_arrs = []
        for x in outputs.values():
            if len(x) >= 2:
                output_arrs.append(x)
        return output_arrs

"""    
[4,7,6,7]

i = 0

outputs_begin_idx_i=[[4]]
j = 1
outputs_begin_idx_i=[
    [4],
    [4,7]
]

j = 2
nums[j] = 6
outputs_begin_idx_i=[
    [4], [4, 6]
    [4,7]
]

j = 3
nums[j] = 7
outputs_begin_idx_i=[
    [4], 
    [4,7],
    [4,6],

    [4,7],
    [4,7,7],
    [4,6,7]
]

"""

26200