<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]]:
        res = []
        memo = defaultdict(int)
        n = len(nums)
        stack = [([nums[i]], i) for i in range(n - 1)]
        while stack:
            candidate, idx = stack.pop(0)
            if len(candidate) >= 2:
                res.append(candidate)
            for i in range(idx + 1, n):
                if nums[i] >= nums[idx]:
                    next_candidate = candidate + [nums[i]]
                    next_candidate_hashed = f'{next_candidate}'
                    if memo[next_candidate_hashed] == 0:
                        memo[next_candidate_hashed] = 1
                        stack.append((next_candidate, i))
        return res



            
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