第一个数组中的元素计数大于第二个数组，每个元素只考虑一次

给定两个大小为N 的排序数组。任务是找到第一个数组中严格大于第二个数组元素的最大元素数，使得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：

1. 一一比较索引为 0的两个数组的元素。
2. 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。
3. 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。
4. 重复上述步骤，直到任何数组的索引到达最后一个元素。
5. 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

Javascript

 
// Javascript program for the above approach
 
// Function to find greater elements
function findMaxElements( arr1, arr2, n)
{
    // Index counter for arr1
    var cnt1 = 0;
     
    // Index counter for arr2
    var cnt2 = 0;
     
    // To store the maximum elements
    var 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
    document.write( maxelements );
}
 
var arr1 = [10, 15, 20, 25, 30, 35];
var arr2 = [12, 14, 26, 32, 34, 40];
var n = arr1.length;
findMaxElements(arr1, arr2, n);
 
// This code is contributed by rrrtnx.

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)

如果您想与行业专家一起参加直播课程，请参阅Geeks Classes Live