<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]]:

        self.num = nums
        res = self.bt(len(nums) - 1, 100)
        res = list(set([tuple(_) for _ in res if len(_) > 1]))
        return [list(_) for _ in res]


    def bt(self, loc: int, max_end: int):

        if not loc: return [] if self.num[0] > max_end else [[self.num[0]]]
        if self.num[loc] > max_end:
            return self.bt(loc - 1, max_end)
        else:
            pre = self.bt(loc - 1, max_end)
            temp = self.bt(loc - 1, self.num[loc])
            if temp: 
                temp = [_ + [self.num[loc]] for _ in temp]
                temp += [[self.num[loc]]]
            else:
                temp = [[self.num[loc]]]
            return temp + pre

        
28000