Given two binary strings A and B of equal lengths, the task is to print a string that is the XOR of Binary Strings A and B.
Examples:
Input: A = "0001", B = "0010"
Output: 0011Input: A = "1010", B = "0101"
Output: 1111
Approach: The idea is to iterate over both the string character by character and if the character is mismatched then add "1" as the character in the answer string otherwise add "0" to the answer string to generate the XOR string.
Below is the implementation of the above approach:
// C++ Implementation to find the
// XOR of the two Binary Strings
#include<bits/stdc++.h>
using namespace std;
// Function to find the
// XOR of the two Binary Strings
string xoring(string a, string b, int n){
string ans = "";
// Loop to iterate over the
// Binary Strings
for (int i = 0; i < n; i++)
{
// If the Character matches
if (a[i] == b[i])
ans += "0";
else
ans += "1";
}
return ans;
}
// Driver Code
int main()
{
string a = "1010";
string b = "1101";
int n = a.length();
string c = xoring(a, b, n);
cout << c << endl;
}
// This code is contributed by Surendra_Gangwar
// Java Implementation to find the
// XOR of the two Binary Strings
import java.io.*;
class GFG {
// Function to find the
// XOR of the two Binary Strings
static String xoring(String a, String b, int n){
String ans = "";
// Loop to iterate over the
// Binary Strings
for (int i = 0; i < n; i++)
{
// If the Character matches
if (a.charAt(i) == b.charAt(i))
ans += "0";
else
ans += "1";
}
return ans;
}
// Driver Code
public static void main (String[] args)
{
String a = "1010";
String b = "1101";
int n = a.length();
String c = xoring(a, b, n);
System.out.println(c);
}
}
// This code is contributed by shubhamsingh10
# Python Implementation to find the
# XOR of the two Binary Strings
# Function to find the
# XOR of the two Binary Strings
def xor(a, b, n):
ans = ""
# Loop to iterate over the
# Binary Strings
for i in range(n):
# If the Character matches
if (a[i] == b[i]):
ans += "0"
else:
ans += "1"
return ans
# Driver Code
if __name__ == "__main__":
a = "1010"
b = "1101"
n = len(a)
c = xor(a, b, n)
print(c)
// C# Implementation to find the
// XOR of the two Binary Strings
using System;
class GFG{
// Function to find the
// XOR of the two Binary Strings
static string xoring(string a, string b, int n){
string ans = "";
// Loop to iterate over the
// Binary Strings
for (int i = 0; i < n; i++)
{
// If the Character matches
if (a[i] == b[i])
ans += "0";
else
ans += "1";
}
return ans;
}
// Driver Code
static public void Main ()
{
string a = "1010";
string b = "1101";
int n = a.Length;
string c = xoring(a, b, n);
Console.WriteLine(c);
}
}
// This code is contributed by shubhamsingh10
<script>
// Javascript Implementation to find the
// XOR of the two Binary Strings
// Function to find the
// XOR of the two Binary Strings
function xoring(a, b, n){
let ans = "";
// Loop to iterate over the
// Binary Strings
for (let i = 0; i < n; i++)
{
// If the Character matches
if (a[i] == b[i])
ans += "0";
else
ans += "1";
}
return ans;
}
// Driver Code
let a = "1010";
let b = "1101";
let n = a.length;
let c = xoring(a, b, n);
document.write(c);
</script>
Output
0111
Time Complexity: O(N)
Auxiliary Space: O(N)
method2:performs XOR operation bit by bit to generate the XOR result string.
Approach: We take two binary strings A and B as input which are of equal length, and apply for loop and performs XOR operation bit by bit to generate the XOR result. Then we print the XOR result string.
Below is the implementation of the above approach:
#include <bits/stdc++.h>
using namespace std;
// Function to perform XOR of two binary strings
std::string xor_Binary_Strings( string &A,string &B) {
int len = A.length(); // Get the length of binary strings A and B
string result = ""; // Initialize the result string
// Perform XOR operation bit by bit
for (int i = 0; i < len; i++) {
// XOR of current bits and convert it to char ('0' or '1')
char currentXOR = ((A[i] - '0') ^ (B[i] - '0')) + '0';
// Append the currentXOR to the result string
result += currentXOR;
}
return result;
}
int main() {
// Input two binary strings of equal length
string A= "1010";
string B = "1101";
int n = A.length();
string C =xor_Binary_Strings(A,B);
cout << C << endl;
return 0;
}
public class Main {
// Function to perform XOR of two binary strings
public static String xorBinaryStrings(String A,
String B)
{
int len = A.length(); // Get the length of binary
// strings A and B
StringBuilder result
= new StringBuilder(); // Initialize the result
// StringBuilder
// Perform XOR operation bit by bit
for (int i = 0; i < len; i++) {
// XOR of current bits and convert it to char
// ('0' or '1')
char currentXOR
= (char)((A.charAt(i) - '0')
^ (B.charAt(i) - '0') + '0');
// Append the currentXOR to the result
// StringBuilder
result.append(currentXOR);
}
return result.toString();
}
public static void main(String[] args)
{
// Input two binary strings of equal length
String A = "1010";
String B = "1101";
int n = A.length();
String C = xorBinaryStrings(A, B);
System.out.println(C);
}
}
// This code is contributed by Samim Hossain Mondal.
# Function to perform XOR of two binary strings
def xor_binary_strings(A, B):
length = len(A) # Get the length of binary strings A and B
result = "" # Initialize the result string
# Perform XOR operation bit by bit
for i in range(length):
# XOR of current bits and convert it to a character ('0' or '1')
current_xor = str(int(A[i]) ^ int(B[i]))
# Append the current_xor to the result string
result += current_xor
return result
def main():
# Input two binary strings of equal length
A = "1010"
B = "1101"
C = xor_binary_strings(A, B)
print(C)
if __name__ == "__main__":
main()
using System;
class Program {
// Function to perform XOR of two binary strings
static string XorBinaryStrings(string A, string B)
{
int len = A.Length; // Get the length of binary
// strings A and B
string result = ""; // Initialize the result string
// Perform XOR operation bit by bit
for (int i = 0; i < len; i++) {
// XOR of current bits and convert it to char
// ('0' or '1')
char currentXOR
= (char)((A[i] - '0') ^ (B[i] - '0') + '0');
// Append the currentXOR to the result string
result += currentXOR;
}
return result;
}
static void Main(string[] args)
{
// Input two binary strings of equal length
string A = "1010";
string B = "1101";
int n = A.Length;
string C = XorBinaryStrings(A, B);
Console.WriteLine(C);
}
}
// Function to perform XOR of two binary strings
function xorBinaryStrings(A, B) {
const len = A.length; // Get the length of binary strings A and B
let result = ''; // Initialize the result string
// Perform XOR operation bit by bit
for (let i = 0; i < len; i++) {
// XOR of current bits and convert it to '0' or '1'
const currentXOR = String(Number(A[i]) ^ Number(B[i]));
// Append the currentXOR to the result string
result += currentXOR;
}
return result;
}
// Main function
function main() {
// Input two binary strings of equal length
const A = '1010';
const B = '1101';
const C = xorBinaryStrings(A, B);
console.log(C);
}
// Call the main function
main();
Output
0111
Time Complexity: O(N)
Auxiliary Space: O(N)
