Character pairs from two strings with even sum

Given two strings s1 and s2. The task is to take one character from first string, and one character from second string, and check if the sum of ascii values of both the character is an even number. Print the total number of such pairs. Note that both the strings consist of lowercase English alphabets.
Examples:

Input: s1 = “geeks”, s2 = “for” 
Output: 5 
All valid pairs are: 
(g, o) -> 103 + 111 = 214 
(e, o) -> 101 + 111 = 212 
(e, o) -> 101 + 111 = 212 
(k, o) -> 107 + 111 = 218 
(s, o) -> 115 + 111 = 226
Input: s1 = “abcd”, s2 = “swed” 
Output: 8 

Approach:

• It is clear, that for the sum to be even, either both the ascii values must be even or both must be odd.
• Calculate the total number of odd and even ascii values from first string. Let it be a1 and b1 respectively.
• Calculate the total number of odd and even ascii values from second string. Let it be a2 and b2 respectively.
• Then the total number of valid pairs will be ((a1 * a2) + (b1 * b2)).

Below is the implementation of the above approach: 

C++

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

// Function to return the total number of valid pairs
int totalPairs(string s1, string s2)
{
    int a1 = 0, b1 = 0;

    // Count total number of even and odd
    // ascii values for string s1
    for (int i = 0; i < s1.length(); i++) {
        if (int(s1[i]) % 2 != 0)
            a1++;
        else
            b1++;
    }

    int a2 = 0, b2 = 0;

    // Count total number of even and odd
    // ascii values for string s2
    for (int i = 0; i < s2.length(); i++) {
        if (int(s2[i]) % 2 != 0)
            a2++;
        else
            b2++;
    }

    // Return total valid pairs
    return ((a1 * a2) + (b1 * b2));
}

// Driver code
int main()
{
    string s1 = "geeks", s2 = "for";
    cout << totalPairs(s1, s2);

    return 0;
}

Java

// Java implementation of the approach
class GfG
{

    // Function to return the total number of valid pairs
    static int totalPairs(String s1, String s2)
    {
        int a1 = 0, b1 = 0;

        // Count total number of even and odd
        // ascii values for string s1
        for (int i = 0; i < s1.length(); i++)
        {
            if ((int)s1.charAt(i) % 2 != 0)
                a1++;
            else
                b1++;
        }

        int a2 = 0, b2 = 0;

        // Count total number of even and odd
        // ascii values for string s2
        for (int i = 0; i < s2.length(); i++)
        {
            if ((int)s2.charAt(i) % 2 != 0)
                a2++;
            else
                b2++;
        }

        // Return total valid pairs
        return ((a1 * a2) + (b1 * b2));
    }

    // Driver code
    public static void main(String []args)
    {

        String s1 = "geeks", s2 = "for";
        System.out.println(totalPairs(s1, s2));
    }
}

// This code is contributed by Rituraj Jain

Python3

# Python3 implementation of the approach

# Function to return the total
# number of valid pairs
def totalPairs(s1, s2) :

    a1 = 0; b1 = 0;

    # Count total number of even and
    # odd ascii values for string s1
    for i in range(len(s1)) :
        if (ord(s1[i]) % 2 != 0) :
            a1 += 1;
        else :
            b1 += 1;

    a2 = 0 ; b2 = 0;

    # Count total number of even and odd
    # ascii values for string s2
    for i in range(len(s2)) :
        if (ord(s2[i]) % 2 != 0) :
            a2 += 1;
        else :
            b2 += 1;

    # Return total valid pairs
    return ((a1 * a2) + (b1 * b2));

# Driver code
if __name__ == "__main__" :

    s1 = "geeks";
    s2 = "for";

    print(totalPairs(s1, s2));

# This code is contributed by Ryuga

C#

// C# implementation of the approach
using System;

class GfG
{

    // Function to return the total number of valid pairs
    static int totalPairs(String s1, String s2)
    {
        int a1 = 0, b1 = 0;

        // Count total number of even and odd
        // ascii values for string s1
        for (int i = 0; i < s1.Length; i++)
        {
            if ((int)s1[i] % 2 != 0)
                a1++;
            else
                b1++;
        }

        int a2 = 0, b2 = 0;

        // Count total number of even and odd
        // ascii values for string s2
        for (int i = 0; i < s2.Length; i++)
        {
            if ((int)s2[i] % 2 != 0)
                a2++;
            else
                b2++;
        }

        // Return total valid pairs
        return ((a1 * a2) + (b1 * b2));
    }

    // Driver code
    public static void Main(String []args)
    {

        String s1 = "geeks", s2 = "for";
        Console.WriteLine(totalPairs(s1, s2));
    }
}

// This code is contributed by 29AjayKumar

Javascript

<script>

// Javascript implementation of the approach

// Function to return the total number of valid pairs
function totalPairs(s1, s2)
{
    var a1 = 0, b1 = 0;

    // Count total number of even and odd
    // ascii values for string s1
    for (var i = 0; i < s1.length; i++) {
        if ((s1[i].charCodeAt(0)) % 2 != 0)
            a1++;
        else
            b1++;
    }

    var a2 = 0, b2 = 0;

    // Count total number of even and odd
    // ascii values for string s2
    for (var i = 0; i < s2.length; i++) {
        if ((s2[i].charCodeAt(0)) % 2 != 0)
            a2++;
        else
            b2++;
    }

    // Return total valid pairs
    return ((a1 * a2) + (b1 * b2));
}

// Driver code
var s1 = "geeks", s2 = "for";
document.write( totalPairs(s1, s2));

</script>

Output:

5

Time Complexity: O(m + n), where m and n are the lengths of the two strings.
Auxiliary Space: O(1), no extra space is required, so it is a constant.