<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]]:
        ans = []
        seen =set()
        def backtracking(start_idx, curr_res):
            res_str = ",".join(str(x) for x in curr_res)
            if res_str in seen:
                return
            if len(curr_res) > 1:
                if curr_res[-1] - curr_res[-2] >= 0:
                    ans.append(curr_res[:])
                    seen.add(res_str)
                else:
                    return
            for i in range(start_idx, len(nums)):
                curr_res.append(nums[i])
                backtracking(i+1, curr_res)
                curr_res.pop()
        backtracking(0, [])
        return ans
28600