Skip to content
Related Articles

Related Articles

Count Pairs from two arrays with even sum

Improve Article
Save Article
  • Last Updated : 31 May, 2022
Improve Article
Save Article

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:
{9, 7}, {14, 8}, {6, 4} and {2, 20} are the valid pairs.
Input: A[] = {2, 4, 6}, B[] = {8, 10, 12} 
Output:
 

 

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++




// 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;
}

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..

PHP




<?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
?>

Javascript




<script>
 
// Javascript 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
        var odd1 = 0, even1 = 0;
        var 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
        var pairs = Math.min(odd1, odd2) +
                    Math.min(even1, even2);
 
        // Return the number of pairs
        return pairs;
    }
 
    // Driver code
     
 
        var a = [ 9, 14, 6, 2, 11 ];
        var b = [ 8, 4, 7, 20 ];
        var n = a.length;
        var m = b.length;
        document.write(count_pairs(a, b, n, m));
 
 
// This code contributed by umadevi9616
 
</script>

Output: 

4

 

Time Complexity: O(n + m)

Auxiliary Space: O(1)


My Personal Notes arrow_drop_up
Related Articles

Start Your Coding Journey Now!