如何比较时间复杂度小于O（n ^ 2）的两个数组中的每个元素 [英] How to compare each element in two arrays with time complexity less than O(n^2)

问题描述

假设我们有两个数组A [n]和b [n​​]，目标是将A中的每个元素与B中的元素进行比较。然后返回一个列表结果[n]，该结果记录A中每个元素的数量，

Suppose we have two arrays A[n] and b[n], the goal is to compare every element in A to elements in B. Then return a list result[n] that records the number of each element in A that is larger than the elements in B.

例如，

A = [38，24，43，3]，B = [9，82，10，11]

A = [38, 24, 43, 3], B = [9, 82, 10, 11]

因为38大于9 ，10和11，所以result [0]为3。
然后结果为[3、3、3、0]。

Since 38 is larger than 9, 10 and 11, so result[0] is 3. Then result is [3, 3, 3, 0].

这将是最好的

谢谢。

推荐答案

您可以以O（nlogn）复杂度执行上述算法，其中n是问题中给出的数组A和数组B的长度。

You can perform the above algorithm in O(nlogn) complexity where n is the length of array A and array B as given in the question.

1. Sort both the arrays A and B, this will take O(nlogn) time complexity.
2. Take two pointers i and j, initialize both of them to 0. we will use i for array A and j for B.
3. Create a result array res of size n.
4. Start a while loop 
   while(i<n && j<n) {
     if(A[i] > B[j]) {
       j++;
     } else {
       res[i] = j+1;
       i++;
     }
   }
5. while(i<n) {
     res[i] = n;
   }
   This step is for the case where all elements in A are bigger than all elements in B.

最后，您将准备好 res 个带有答案的数组。

At the end you will have res array ready with the answer.

总时间复杂性- O（nlogn）。

希望这会有所帮助！

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