该组合算法的时间复杂度 [英] Time complexity of this combination algorithms

问题描述

我正在实现一种组合算法.这将从给定列表中生成长度的唯一组合.

I was implementing a combinations algorithm. This should generate unique combinations of length from the given list .

例如:

Input list [1, 2, 3, 4, 5] with k = 3
should generate output

    [1, 2, 3]
    [1, 2, 4]
    [1, 2, 5]
    [1, 3, 4]
    [1, 3, 5]
    [1, 4, 5]
    [2, 3, 4]
    [2, 3, 5]
    [2, 4, 5]
    [3, 4, 5]

下面给出了有效的python代码以供参考.

The working python code is given below for reference.

def my_combinations(items, k, out):
    if k==0:
        print out
        return

    for i in range(len(items)):
        new_out = out[:]
        new_out.append(items[i])
        my_combinations(items[i+1:], k-1, new_out)

问题:

此算法的时间复杂度是多少?

What is the time complexity of this algorithm?

我从递归方程开始.

Base case: T(n, 0) = 1
Recurion : T(n, k) = T(n-1, k-1) + T(n-2, k-1) + T(n-3, k-1) + .. + T(0, k-1) + 1
                   = n * T(n-1, k-1) + 1

T(n)= ???

T(n) = ???

扩展解决方案.

此问题与生成所有组合时的复杂性不同.

我的问题是关于给定实现的时间复杂性，链接问题是关于生成所有组合的运行时间的一般问题.

My question is about the time complexity of the given implementation and the link question talks about runtime of generating all combinations in general.

推荐答案

感谢@Michael Foukarakis指出了丢失的K.

Thanks @Michael Foukarakis for pointing the missing K.

Base case: T(n, 0) = 1
Recurion : T(n, k) = T(n-1, k-1) + T(n-2, k-1) + T(n-3, k-1) + .. + T(0, k-1) + 1
                   = n * T(n-1, k-1) + 1

如下扩展

T(n, k) = n * T(n-1, k-1) + 1
        = n * (n-1) * T(n-2, k-2) + 1 + 1
        = n * (n-1) * T(n-2, k-2) + 2
        = n * (n-1) * (n-2) * T(n-3, k-3) + 3

        ...

         = n * (n-1) * (n-2) * ..(n-k) T(n-k, k-k) + k
         = n * (n-1) * (n-2) * ..(n-k) (1) + k
         = O(n^k)  (As it is a k th order polynomial)

总体而言，我们可以说 O(n ^k )运行时.

Overall we can say O(n^k) runtime.

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