给定两个大小为N的排序数组。任务是在第一个数组中找到最大数量的元素,这些元素严格大于第二个数组中的元素,这样一个元素只能被考虑一次。
例子:
Input: arr1[] = { 20, 30, 50 }, arr2[] = { 25, 40, 60 }
Output: 2
Explanation:
Maximum 2 elements 30 (30 > 25) and 50 (50 > 40) of array arr1 is grater than arr2.
Input: arr1[] = { 10, 15, 20, 25, 30, 35 }, arr2[] = { 12, 14, 26, 32, 34, 40 }
Output: 4
Explanation:
Maximum 4 elements 15 (15 > 12), 20 (20 > 14), 30 (30 > 26) and 35 (35 > 34) of arr1 is grater than arr2.
方法:
- 从索引0开始一个一个地比较两个数组的元素。
- 如果ARR1的索引处的元素ARR2的指标处比元件大于增加的答案和两个阵列的由1的索引。
- 如果ARR1的索引处的元素是更小或ARR2的索引以等于元件然后
增加arr1的索引。 - 重复上述步骤,直到任何数组的索引都到达最后一个元素。
- 打印答案
下面是上述方法的实现:
C++
// C++ program for the above approach
#include
using namespace std;
// Function to find greater elements
void findMaxElements(
int arr1[], int arr2[], int n)
{
// Index counter for arr1
int cnt1 = 0;
// Index counter for arr2
int cnt2 = 0;
// To store the maximum elements
int maxelements = 0;
while (cnt1 < n && cnt2 < n) {
// If element is greater,
// update maxelements and counters
// for both the arrays
if (arr1[cnt1] > arr2[cnt2]) {
maxelements++;
cnt1++;
cnt2++;
}
else {
cnt1++;
}
}
// Print the maximum elements
cout << maxelements << endl;
}
int main()
{
int arr1[] = { 10, 15, 20, 25, 30, 35 };
int arr2[] = { 12, 14, 26, 32, 34, 40 };
int n = sizeof(arr1) / sizeof(arr1[0]);
findMaxElements(arr1, arr2, n);
return 0;
}
Java
// Java program for the above approach
class Main{
// Function to find greater elements
static void findmaxelements(int arr1[], int arr2[], int n)
{
// Index counter for arr1
int cnt1 = 0;
// Index counter for arr1
int cnt2 = 0;
// To store the maximum elements
int maxelements = 0;
while(cnt1 < n && cnt2 < n)
{
// If element is greater,
// update maxelements and counters
// for both the arrays
if(arr1[cnt1] > arr2[cnt2])
{
maxelements++;
cnt1++;
cnt2++;
}
else
{
cnt1++;
}
}
// Print the maximum elements
System.out.println(maxelements);
}
// Driver Code
public static void main(String[] args)
{
int arr1[] = { 10, 15, 20, 25, 30, 35 };
int arr2[] = { 12, 14, 26, 32, 34, 40 };
findmaxelements(arr1, arr2, arr1.length);
}
}
// This code is contributed by divyeshrabadiya07
Python3
# Python3 program for the above approach
# Function to find greater elements
def findmaxelements(arr1, arr2, n):
# Index counter for arr1
cnt1 = 0
# Index counter for arr2
cnt2 = 0
# To store the maximum elements
maxelements = 0
# If element is greater,
# update maxelements and counters
# for both the arrays
while cnt1 < n and cnt2 < n :
if arr1[cnt1] > arr2[cnt2] :
maxelements += 1
cnt1 += 1
cnt2 += 1
else :
cnt1 += 1
# Print the maximum elements
print(maxelements)
# Driver Code
arr1 = [ 10, 15, 20, 25, 30, 35 ]
arr2 = [ 12, 14, 26, 32, 34, 40 ]
findmaxelements(arr1, arr2, len(arr1))
# This code is contributed by divyeshrabadiya07
C#
// C# program for the above approach
using System;
class GFG{
// Function to find greater elements
static void findmaxelements(int[] arr1,
int[] arr2, int n)
{
// Index counter for arr1
int cnt1 = 0;
// Index counter for arr1
int cnt2 = 0;
// To store the maximum elements
int maxelements = 0;
while(cnt1 < n && cnt2 < n)
{
// If element is greater, update
// maxelements and counters for
// both the arrays
if(arr1[cnt1] > arr2[cnt2])
{
maxelements++;
cnt1++;
cnt2++;
}
else
{
cnt1++;
}
}
// Print the maximum elements
Console.Write(maxelements);
}
// Driver Code
static public void Main(string[] args)
{
int[] arr1 = { 10, 15, 20, 25, 30, 35 };
int[] arr2 = { 12, 14, 26, 32, 34, 40 };
findmaxelements(arr1, arr2, arr1.Length);
}
}
// This code is contributed by rutvik_56
输出:
4
时间复杂度: O(N) ,其中N是数组的长度。
空间复杂度: O(1)
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