<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]]:
        # backtracking - every element, include or not
        # we can only include if the item is greater or equal to the last element
        # once we reach end of the nums, append cur to ans if size is at least 2
        def backtrack(nums, cur, idx, ans): # O(2^N * N)
            if len(cur) >= 2:
                ans.add(tuple(cur)) # O(N) time and space
            for i in range(idx, len(nums)):
                if not cur or cur[-1] <= nums[i]:
                    cur.append(nums[i])
                    backtrack(nums, cur, i+1, ans)
                    cur.pop()


        ans = set()
        backtrack(nums, [], 0, ans)
        return [list(l) for l in ans]
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