从两个具有偶数和的数组中计算对

给定两个分别由N和M个整数组成的数组A []和B [] 。任务是计算通过从数组A []中选择一个元素以及从数组B []中选择另一个元素而形成的无序对的数量，以使它们的总和为偶数。
请注意，一个元素将仅是一对的一部分。

例子：

Input: A[] = {9, 14, 6, 2, 11}, B[] = {8, 4, 7, 20}
Output: 4
{9, 7}, {14, 8}, {6, 4} and {2, 20} are the valid pairs.

Input: A[] = {2, 4, 6}, B[] = {8, 10, 12}
Output: 3

方法：对两个数组中的奇数和偶数进行计数，对数的答案将是min(odd1，odd2)+ min(even1，even2)，因为(odd + odd)=偶数和(even +偶数) )=偶数。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
  
// Function to return count of required pairs
int count_pairs(int a[], int b[], int n, int m)
{
  
    // Count of odd and even numbers
    // from both the arrays
    int odd1 = 0, even1 = 0;
    int odd2 = 0, even2 = 0;
  
    // Find the count of odd and
    // even elements in a[]
    for (int i = 0; i < n; i++) {
        if (a[i] % 2 == 1)
            odd1++;
        else
            even1++;
    }
  
    // Find the count of odd and
    // even elements in b[]
    for (int i = 0; i < m; i++) {
        if (b[i] % 2 == 1)
            odd2++;
        else
            even2++;
    }
  
    // Count the number of pairs
    int pairs = min(odd1, odd2) + min(even1, even2);
  
    // Return the number of pairs
    return pairs;
}
  
// Driver code
int main()
{
    int a[] = { 9, 14, 6, 2, 11 };
    int b[] = { 8, 4, 7, 20 };
    int n = sizeof(a) / sizeof(a[0]);
    int m = sizeof(b) / sizeof(b[0]);
    cout << count_pairs(a, b, n, m);
  
    return 0;
}

Java

// Java implementation of the approach
import java.io.*;
  
class GFG
{
  
// Function to return count of required pairs
static int count_pairs(int a[], int b[], int n, int m)
{
  
    // Count of odd and even numbers
    // from both the arrays
    int odd1 = 0, even1 = 0;
    int odd2 = 0, even2 = 0;
  
    // Find the count of odd and
    // even elements in a[]
    for (int i = 0; i < n; i++) 
    {
        if (a[i] % 2 == 1)
            odd1++;
        else
            even1++;
    }
  
    // Find the count of odd and
    // even elements in b[]
    for (int i = 0; i < m; i++)
    {
        if (b[i] % 2 == 1)
            odd2++;
        else
            even2++;
    }
  
    // Count the number of pairs
    int pairs = Math.min(odd1, odd2) + Math.min(even1, even2);
  
    // Return the number of pairs
    return pairs;
}
  
// Driver code
public static void main (String[] args) 
{
      
    int a[] = { 9, 14, 6, 2, 11 };
    int b[] = { 8, 4, 7, 20 };
    int n = a.length;
    int m = b.length;
    System.out.println (count_pairs(a, b, n, m));
  
}
}
  
// This code is contributes by ajit

Python3

# Python 3 implementation of the approach
  
# Function to return count of required pairs
def count_pairs(a,b,n,m):
      
    # Count of odd and even numbers
    # from both the arrays
      
    odd1 = 0
    even1 = 0
    odd2 = 0
    even2 = 0
      
    # Find the count of odd and
    # even elements in a[]
    for i in range(n):
        if (a[i] % 2 == 1):
            odd1 += 1
        else:
            even1 += 1
  
    # Find the count of odd and
    # even elements in b[]
      
    for i in range(m):
        if (b[i] % 2 == 1):
            odd2 += 1
        else:
            even2 += 1
              
    # Count the number of pairs
    pairs = min(odd1, odd2) + min(even1, even2)
      
    # Return the number of pairs
    return pairs
  
# Driver code
if __name__ == '__main__':
    a = [9, 14, 6, 2, 11]
    b = [8, 4, 7, 20]
    n = len(a)
    m = len(b)
    print(count_pairs(a, b, n, m))
  
# This code is contributed by
# Surendra_Gangwar

C#

// C# implementation of the approach
using System;
  
class GFG
{
  
// Function to return count of required pairs
static int count_pairs(int []a, int []b, int n, int m)
{
  
    // Count of odd and even numbers
    // from both the arrays
    int odd1 = 0, even1 = 0;
    int odd2 = 0, even2 = 0;
  
    // Find the count of odd and
    // even elements in a[]
    for (int i = 0; i < n; i++) 
    {
        if (a[i] % 2 == 1)
            odd1++;
        else
            even1++;
    }
  
    // Find the count of odd and
    // even elements in b[]
    for (int i = 0; i < m; i++)
    {
        if (b[i] % 2 == 1)
            odd2++;
        else
            even2++;
    }
  
    // Count the number of pairs
    int pairs = Math.Min(odd1, odd2) + Math.Min(even1, even2);
  
    // Return the number of pairs
    return pairs;
}
  
// Driver code
public static void Main () 
{
      
    int []a = { 9, 14, 6, 2, 11 };
    int []b = { 8, 4, 7, 20 };
    int n = a.Length;
    int m = b.Length;
    Console.WriteLine (count_pairs(a, b, n, m));
  
}
}
  
// This code is contributes by anuj_67..

输出：

4