# Character pairs from two strings with even sum

• Last Updated : 15 Dec, 2022

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:
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:

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 ``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

 ``

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.

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