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 = “neveropen”, 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
Brute Force Approach:
A brute force approach to solve this problem would be to iterate through every possible pair of characters from both the strings, calculate the sum of their ASCII values and check if the sum is even or not. If it is even, then increment a counter. Finally, return the counter as the total number of valid pairs.
Here are the steps to implement above approach:
- Initialize a counter to 0 to keep track of the total number of valid pairs.
- Iterate over the first string s1 and for each character, iterate over the second string s2 and calculate the sum of the ASCII values of both characters.
- Check if the sum of the ASCII values is even or not using the modulo operator (%).
- Return the counter as the total number of valid pairs.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h> using namespace std; // Function to return the total number of valid pairs int totalPairs(string s1, string s2) { int cnt = 0; for ( int i = 0; i < s1.length(); i++) { for ( int j = 0; j < s2.length(); j++) { int sum = s1[i] + s2[j]; if (sum % 2 == 0) cnt++; } } return cnt; } // Driver code int main() { string s1 = "neveropen" , s2 = "for" ; cout << totalPairs(s1, s2); return 0; } |
Java
import java.util.*; public class Main { // Function to return the total number of valid pairs public static int totalPairs(String s1, String s2) { int cnt = 0 ; for ( int i = 0 ; i < s1.length(); i++) { for ( int j = 0 ; j < s2.length(); j++) { int sum = s1.charAt(i) + s2.charAt(j); if (sum % 2 == 0 ) cnt++; } } return cnt; } // Driver code public static void main(String[] args) { String s1 = "neveropen" ; String s2 = "for" ; System.out.println(totalPairs(s1, s2)); } } |
Python3
def totalPairs(s1, s2): cnt = 0 # Initialize a counter to keep track of valid pairs # Nested loops to iterate through each character in both strings for i in range ( len (s1)): for j in range ( len (s2)): _sum = ord (s1[i]) + ord (s2[j]) # Calculate the sum of ASCII values of characters if _sum % 2 = = 0 : # Check if the sum is even cnt + = 1 # Increment the counter for valid pairs return cnt # Return the total count # Driver code if __name__ = = "__main__" : s1 = "neveropen" s2 = "for" print (totalPairs(s1, s2)) |
C#
using System; public class Program { // Function to return the total number of valid pairs static int TotalPairs( string s1, string s2) { int cnt = 0; for ( int i = 0; i < s1.Length; i++) { for ( int j = 0; j < s2.Length; j++) { int sum = s1[i] + s2[j]; if (sum % 2 == 0) cnt++; } } return cnt; } // Driver code public static void Main() { string s1 = "neveropen" , s2 = "for" ; Console.WriteLine(TotalPairs(s1, s2)); } } |
Javascript
// Function to return the total number of valid pairs function totalPairs(s1, s2) { let cnt = 0; // Loop through the characters of s1 for (let i = 0; i < s1.length; i++) { // Loop through the characters of s2 for (let j = 0; j < s2.length; j++) { // Calculate the sum of the ASCII values of characters from s1 and s2 let sum = s1.charCodeAt(i) + s2.charCodeAt(j); // Check if the sum is even if (sum % 2 === 0) { cnt++; } } } return cnt; } // Driver code const s1 = "neveropen" ; const s2 = "for" ; console.log(totalPairs(s1, s2)); |
5
Time Complexity: O(m * n), where m and n are the lengths of the strings s1 and s2, respectively.
Auxiliary Space: O(1)
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 = "neveropen" , 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 = "neveropen" , 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 = "neveropen" ; 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 = "neveropen" , 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 = "neveropen" , s2 = "for" ; document.write( totalPairs(s1, s2)); </script> |
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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