Count Pairs from two arrays with even sum

Given two arrays A[] and B[] of N and M integers respectively. The task is to count the number of unordered pairs formed by choosing an element from array A[] and other from array B[] in such a way that their sum is an even number.
Note that an element will only be a part of a single pair.

Examples:

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



Approach: Count the number of odd and even numbers in both the arrays and the answer to the number of pairs will be min(odd1, odd2) + min(even1, even2) because (odd + odd) = even and (even + even) = even.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
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;
}

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Java

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// 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 

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Python3

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# 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

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C#

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// 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..

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PHP

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<?php
// PHP implementation of the approach 
  
// Function to return count of required pairs 
function 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 = 0; $i < $n; $i++)
    
        if ($a[$i] % 2 == 1) 
            $odd1++; 
        else
            $even1++; 
    
  
    // Find the count of odd and 
    // even elements in b[] 
    for ($i = 0; $i < $m; $i++) 
    
        if ($b[$i] % 2 == 1) 
            $odd2++; 
        else
            $even2++; 
    
  
    // Count the number of pairs 
    $pairs = min($odd1, $odd2) + min($even1, $even2); 
  
    // Return the number of pairs 
    return $pairs
  
// Driver code 
$a = array( 9, 14, 6, 2, 11 ); 
$b = array( 8, 4, 7, 20 ); 
$n = count($a); 
$m = count($b); 
  
echo count_pairs($a, $b, $n, $m); 
  
// This code is contributes by AnkitRai01
?>

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Output:

4


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Striver(underscore)79 at Codechef and codeforces D

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