Given 2 int arrays, a and b, each length 3, return a new array length 2 containing their middle elements.

以下是 Python 代码实现：

def middle_way(a, b):
    return [a[1], b[1]]

这个函数接受两个长度为 3 的整数数组 `a` 和 `b` 作为输入，返回一个新的长度为 2 的数组，其中包含两个输入数组的中间元素。

例如，`middle_way([1, 2, 3], [4, 5, 6])` 的返回值为 `[2, 5]`。

相关问题

int []arr3=new int[arr1.length]

这行代码创建了一个名为 arr3 的整型数组，其长度与 arr1 数组的长度相同。换句话说，arr3 数组的容量与 arr1 数组相同，但是它们是两个不同的数组对象。该代码还没有对 arr3 数组进行初始化，因此数组中的元素将被默认初始化为 0。如果需要对 arr3 数组进行初始化，可以使用循环或者 Arrays.fill() 方法。

You are given two arrays a and b , both of length n . Your task is to count the number of pairs of integers (i,j) such that 1≤i<j≤n and ai⋅aj=bi+bj .

To solve this problem, we can first preprocess the array b to create a hash table that maps each value in b to its frequency. Then, for each pair of indices (i,j) such that i<j, we can check if ai⋅aj is equal to bi+bj by computing ai/bj and aj/bi and checking if they are equal. If they are equal, we can check if the values ai/bj and aj/bi are present in the hash table. If both values are present, we can increment our count by the product of their frequencies in the hash table.

The time complexity of this algorithm is O(n^2) due to the nested loops, but the constant factor is small due to the hash table lookups. Here is the Python code for the algorithm:

from collections import defaultdict

def count_pairs(a, b):
    n = len(a)
    freq = defaultdict(int)
    for x in b:
        freq[x] += 1
    count = 0
    for i in range(n):
        for j in range(i+1, n):
            if a[i]*a[j] == b[i]+b[j]:
                if b[i] != 0 and b[j] % b[i] == 0 and a[j] == (b[j] // b[i]) * a[i]:
                    x = b[i]
                    y = b[j] // b[i]
                    if x in freq and y in freq:
                        count += freq[x] * freq[y]
                if b[j] != 0 and b[i] % b[j] == 0 and a[i] == (b[i] // b[j]) * a[j]:
                    x = b[j]
                    y = b[i] // b[j]
                    if x in freq and y in freq:
                        count += freq[x] * freq[y]
    return count

Note that we need to check both possible orderings of ai and aj in case one of them is zero or the corresponding value in b is not present in the hash table.