计数不同的可能对，以使A中的元素大于B中的元素

给定两个给定的相等长度的数组A和B ，任务是找到可以选择的最大对不同元素对，以使A中的元素严格大于B中的元素。

例子：

Input:
A[]={20, 30, 50} , B[]={25, 60, 40}
Output: 2
Explanation:
(30, 25) and (50, 40) are the two possible pairs.

Input:
A[]={20, 25, 60} , B[]={25, 60, 40}
Output: 1
Explanation:
(60, 25) or (60, 40) can be the required pair.

方法：为了解决此问题，我们需要采用以下方法：

• 对两个数组进行排序。
• 遍历数组A的整个长度，并检查A中的当前元素是否大于B中当前指向B的元素。
• 如果是这样，则指向两个数组中的下一个元素。否则，移至A中的下一个元素，并检查它是否大于当前指向B的元素。
• 在数组A的整个遍历之后，在数组B中遍历的元素数量给出了所需的答案。

Illustration:
Let us consider the two following arrays:
A[] = { 30, 28, 45, 22 } , B[] = { 35, 25, 22, 48 }
After sorting, the arrays appear to be
A[] = { 22, 28, 30, 45 } , B[] = { 22, 25, 35, 48}

After the first iteration, since A[0] is not greater than B[0], we move to A[1].
After the second iteration, we move to B[1] as A[1] is greater than B[0].
After the third iteration, we move to B[2] as A[2] is greater than B[1].
Similarly, A[3] is greater than B[2] and we move to B[3].
Hence the number of elements traversed in B,i.e. 3 is the final answer.
The possible pairs are (28,22), (30,25) are (45, 35).

下面是上述方法的实现：

C++

// C++ Program to count number of distinct 
// pairs possible from the two arrays
// such that element selected from one array is 
// always greater than the one selected from 
// the other array
  
#include 
using namespace std;
  
// Function to return the count 
// of pairs
int countPairs(vector A,
                    vector B)
{
    int n = A.size();
      
    sort(A.begin(),A.end());
    sort(B.begin(),B.end());
      
    int ans = 0, i;
    for (int i = 0; i < n; i++) {
        if (A[i] > B[ans]) {
            ans++;
        }
    }
      
    return ans;
}
  
// Driver code
int main()
{
    vector A = { 30, 28, 45, 22 };
    vector B = { 35, 25, 22, 48 };
      
    cout << countPairs(A,B);
      
    return 0;
}

Java

// Java program to count number of distinct 
// pairs possible from the two arrays such 
// that element selected from one array is 
// always greater than the one selected from 
// the other array
import java.util.*;
  
class GFG{
  
// Function to return the count 
// of pairs
static int countPairs(int [] A,
                      int [] B)
{
    int n = A.length;
    int ans = 0;
      
    Arrays.sort(A);
    Arrays.sort(B);
      
    for(int i = 0; i < n; i++)
    {
       if (A[i] > B[ans])
       {
           ans++;
       }
    }
    return ans;
}
  
// Driver code
public static void main(String[] args)
{
    int [] A = { 30, 28, 45, 22 };
    int [] B = { 35, 25, 22, 48 };
      
    System.out.print(countPairs(A, B));
}
}
  
// This code is contributed by sapnasingh4991

Python3

# Python3 program to count number of distinct
# pairs possible from the two arrays
# such that element selected from one array is
# always greater than the one selected from
# the other array
  
# Function to return the count
# of pairs
def countPairs(A, B):
  
    n = len(A)
  
    A.sort()
    B.sort()
  
    ans = 0
    for i in range(n):
        if(A[i] > B[ans]):
            ans += 1
  
    return ans
  
# Driver code
if __name__ == '__main__':
    A = [30, 28, 45, 22]
    B = [35, 25, 22, 48]
  
    print(countPairs(A, B))
  
# This code is contributed by Shivam Singh

C#

// C# program to count number of distinct 
// pairs possible from the two arrays such 
// that element selected from one array is 
// always greater than the one selected from 
// the other array
using System;
class GFG{
  
// Function to return the count 
// of pairs
static int countPairs(int [] A,
                      int [] B)
{
    int n = A.Length;
    int ans = 0;
      
    Array.Sort(A);
    Array.Sort(B);
      
    for(int i = 0; i < n; i++)
    {
        if (A[i] > B[ans])
        {
            ans++;
        }
    }
    return ans;
}
  
// Driver code
public static void Main()
{
    int []A = { 30, 28, 45, 22 };
    int []B = { 35, 25, 22, 48 };
      
    Console.Write(countPairs(A, B));
}
}
  
// This code is contributed by Code_Mech

输出：

3