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Program to find the XOR of ASCII values of characters in a string

Last Updated : 08 Feb, 2024
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Given a string str, the task is to find the XOR of ASCII values of characters in the string.

Examples: 

Input: str = “Geeks” 
Output: 95 
ASCII value of G = 71 
ASCII value of e = 101 
ASCII value of e = 101 
ASCII value of k = 107 
ASCII value of s = 115 
XOR of ASCII values = 71 ^ 101 ^ 101 ^ 107 ^ 115 = 95

Input: str = “GfG” 
Output: 102 

Approach: The idea is to find out the ASCII value of each character one by one and find the XOR value of these values.

Below is the implementation of the above approach: 

C++




// C++ program to find XOR of ASCII
// value of characters in string
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the XOR of ASCII
// value of characters in string
int XorAscii(string str, int len)
{
 
    // store value of first character
    int ans = int(str[0]);
 
    for (int i = 1; i < len; i++) {
 
        // Traverse string to find the XOR
        ans = (ans ^ (int(str[i])));
    }
 
    // Return the XOR
    return ans;
}
 
// Driver code
int main()
{
 
    string str = "geeksforgeeks";
    int len = str.length();
    cout << XorAscii(str, len) << endl;
 
    str = "GfG";
    len = str.length();
    cout << XorAscii(str, len);
 
    return 0;
}


Java




// Java program to find XOR of ASCII
// value of characters in String
class GFG{
  
// Function to find the XOR of ASCII
// value of characters in String
static int XorAscii(String str, int len)
{
  
    // store value of first character
    int ans = (str.charAt(0));
  
    for (int i = 1; i < len; i++) {
  
        // Traverse String to find the XOR
        ans = (ans ^ ((str.charAt(i))));
    }
  
    // Return the XOR
    return ans;
}
  
// Driver code
public static void main(String[] args)
{
  
    String str = "geeksforgeeks";
    int len = str.length();
    System.out.print(XorAscii(str, len) +"\n");
  
    str = "GfG";
    len = str.length();
    System.out.print(XorAscii(str, len));
}
}
 
// This code is contributed by PrinciRaj1992


Python3




# Python3 program to find XOR of ASCII
# value of characters in string
 
# Function to find the XOR of ASCII
# value of characters in string
def XorAscii(str1, len1):
 
    # store value of first character
    ans = ord(str1[0])
 
    for i in range(1,len1):
 
        # Traverse string to find the XOR
        ans = (ans ^ (ord(str1[i])))
 
    # Return the XOR
    return ans
 
# Driver code
str1 = "geeksforgeeks"
len1 = len(str1)
print(XorAscii(str1, len1))
 
str1 = "GfG"
len1 = len(str1)
print(XorAscii(str1, len1))
 
# This code is contributed by mohit kumar 29


C#




// C# program to find XOR of ASCII
// value of characters in String
using System;
 
class GFG{
   
// Function to find the XOR of ASCII
// value of characters in String
static int XorAscii(String str, int len)
{
   
    // store value of first character
    int ans = (str[0]);
   
    for (int i = 1; i < len; i++) {
   
        // Traverse String to find the XOR
        ans = (ans ^ ((str[i])));
    }
   
    // Return the XOR
    return ans;
}
   
// Driver code
public static void Main(String[] args)
{
   
    String str = "geeksforgeeks";
    int len = str.Length;
    Console.Write(XorAscii(str, len) +"\n");
   
    str = "GfG";
    len = str.Length;
    Console.Write(XorAscii(str, len));
}
}
 
// This code is contributed by PrinciRaj1992


Javascript




<script>
 
// Javascript program to find XOR of ASCII
// value of characters in string
 
// Function to find the XOR of ASCII
// value of characters in string
function XorAscii(str, len)
{
 
    // store value of first character
    let ans = str.codePointAt(0);
 
    for (let i = 1; i < len; i++) {
 
        // Traverse string to find the XOR
        ans = (ans ^ (str.codePointAt(i)));
    }
 
    // Return the XOR
    return ans;
}
 
// Driver code
 
    let str = "geeksforgeeks";
    let len = str.length;
    document.write(XorAscii(str, len) + "<br>");
 
    str = "GfG";
    len = str.length;
    document.write(XorAscii(str, len));
 
</script>


Output

123
102

Time Complexity: O(N), where N is the length of string.
Auxiliary Space: O(1), no extra space is required, so it is a constant.



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