<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]]:
        """
        - For each element, backtrack if possible
        """
        res = {}
        self.backtrack(0, nums, [], res)
        # return list(map(
        #     lambda r: list(map(lambda d: int(d), r.split(","))), 
        #     res
        # ))
        return res.values()

    def backtrack(self, i: int, nums: List[int], curr: [], res: dict[str, List[int]]):
        if i >= len(nums):
            curr_key = ",".join(list(map(lambda d: str(d), curr)))
            if len(curr) > 1:
                res[curr_key] = curr.copy()
            return

        self.backtrack(i+1, nums, curr, res)

        if len(curr) == 0 or curr[-1] <= nums[i]:
            curr.append(nums[i])
            self.backtrack(i+1, nums, curr, res)
            curr.pop()
26900