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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 

Output:

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My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

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