是否有一种有效的方法可以仅查找v In Inter P Inter P Inter的感兴趣的组合

发布于 2025-01-29 03:21:18 字数 516 浏览 3 评论 0原文

是否有一种有效的方法只能找到P Inted p Inter P中的V In Imption的组合，

我有类似的列表。

V = [2,3,5,7,9,12]

V的大小可以从0到50＆nbsp;根据我的输入。 V没有重复。

我还有另一个列表p这样的＆nbsp;

P = [ [1],[1,2],[2,3] [3,4], [3,5,7], [3,5,8] ...]

p是恒定的，尺寸为1000。p

没有重复项，在p中，每个元素都按升序排序。＆nbsp;

P中找到V中存在的V组合的一种有效方法是

。

在 BE [[2,3]，[3,5,7]。

我已经尝试从V产生所有可能的安排，然后过滤＆nbsp; by＆nbsp; nbsp;这不是很高效。只是想知道是否有更好的方法可以做我要求的。＆nbsp;

既然我已经知道P是任何形式的树是一个更好的选择吗？

原文

Is there a efficient way to find only the combinations of interest from V exciting in P

I have a list of something like this.

V = [2,3,5,7,9,12]

Size of V can be anywhere from 0 to 50 based on my input.
V has no duplicates.

I have another list P something like this

P = [ [1],[1,2],[2,3] [3,4], [3,5,7], [3,5,8] ...]

P is constant with size 1000.

P has no duplicates and within P, each element is sorted in ascending order.

What would be an efficient way to find the combinations from V existing in P.

For example

When V = [2,3,5,7,9,12] combination of interest would be [[2,3],[3,5,7]].

I have already tried generating all possible arrangements from V and then filter by P. This is not very efficient. Just wondering if there is a better way to do what I asked for.

Since I already know P is any form of Tree a better option for this ?

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难理解 2025-02-05 03:21:18

在实践中，使用set.issuperset可能是最快的方法：

v = set(V)
[c for c in P if v.issuperset(c)]

也可以使用两个指针，沿每个列表逐渐前进。在Python中可能会较慢，但在C/C ++中更快：

def is_subset(V, combination):
    # both V and combination are assumed sorted
    v = iter(V)
    for element in combination:
        for el in v:
            if el >= element:
                break
        if el != element:
            return False
    return True

[c for c in P if is_subset(V, c)]

Using set.issuperset is probably the fastest method in practice:

v = set(V)
[c for c in P if v.issuperset(c)]

It is also possible to use two pointers, gradually advancing along each list. It will probably be slower in Python, but faster in C/C++:

def is_subset(V, combination):
    # both V and combination are assumed sorted
    v = iter(V)
    for element in combination:
        for el in v:
            if el >= element:
                break
        if el != element:
            return False
    return True

[c for c in P if is_subset(V, c)]

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