XOR two binary strings of unequal lengths
Last Updated :
29 Oct, 2023
Given two binary string of unequal lengths A and B, the task is to print the binary string which is the XOR of A and B.
Examples:
Input: A = "11001", B = "111111"
Output: 100110
Input: A = "11111", B = "0"
Output: 11111
Approach: The idea is to first make both the strings of equal length and then perform the XOR of each character one by one and store it in the resultant string.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to insert n 0s in the
// beginning of the given string
void addZeros(string& str, int n)
{
for (int i = 0; i < n; i++) {
str = "0" + str;
}
}
// Function to return the XOR
// of the given strings
string getXOR(string a, string b)
{
// Lengths of the given strings
int aLen = a.length();
int bLen = b.length();
// Make both the strings of equal lengths
// by inserting 0s in the beginning
if (aLen > bLen) {
addZeros(b, aLen - bLen);
}
else if (bLen > aLen) {
addZeros(a, bLen - aLen);
}
// Updated length
int len = max(aLen, bLen);
// To store the resultant XOR
string res = "";
for (int i = 0; i < len; i++) {
if (a[i] == b[i])
res += "0";
else
res += "1";
}
return res;
}
// Driver code
int main()
{
string a = "11001", b = "111111";
cout << getXOR(a, b);
return 0;
}
// Java implementation of the approach
class GFG
{
// Function to insert n 0s in the
// beginning of the given string
static String addZeros(String str, int n)
{
for (int i = 0; i < n; i++)
{
str = "0" + str;
}
return str;
}
// Function to return the XOR
// of the given strings
static String getXOR(String a, String b)
{
// Lengths of the given strings
int aLen = a.length();
int bLen = b.length();
// Make both the strings of equal lengths
// by inserting 0s in the beginning
if (aLen > bLen)
{
a = addZeros(b, aLen - bLen);
}
else if (bLen > aLen)
{
a = addZeros(a, bLen - aLen);
}
// Updated length
int len = Math.max(aLen, bLen);
// To store the resultant XOR
String res = "";
for (int i = 0; i < len; i++)
{
if (a.charAt(i) == b.charAt(i))
res += "0";
else
res += "1";
}
return res;
}
// Driver code
public static void main (String[] args)
{
String a = "11001", b = "111111";
System.out.println(getXOR(a, b));
}
}
// This code is contributed by AnkitRai01
# Python3 implementation of the approach
# Function to insert n 0s in the
# beginning of the given string
def addZeros(strr, n):
for i in range(n):
strr = "0" + strr
return strr
# Function to return the XOR
# of the given strings
def getXOR(a, b):
# Lengths of the given strings
aLen = len(a)
bLen = len(b)
# Make both the strings of equal lengths
# by inserting 0s in the beginning
if (aLen > bLen):
b = addZeros(b, aLen - bLen)
elif (bLen > aLen):
a = addZeros(a, bLen - aLen)
# Updated length
lenn = max(aLen, bLen)
# To store the resultant XOR
res = ""
for i in range(lenn):
if (a[i] == b[i]):
res += "0"
else:
res += "1"
return res
# Driver code
a = "11001"
b = "111111"
print(getXOR(a, b))
# This code is contributed by Mohit Kumar
// C# implementation of the approach
using System;
class GFG
{
// Function to insert n 0s in the
// beginning of the given string
static String addZeros(String str, int n)
{
for (int i = 0; i < n; i++)
{
str = "0" + str;
}
return str;
}
// Function to return the XOR
// of the given strings
static String getXOR(String a, String b)
{
// Lengths of the given strings
int aLen = a.Length;
int bLen = b.Length;
// Make both the strings of equal lengths
// by inserting 0s in the beginning
if (aLen > bLen)
{
a = addZeros(b, aLen - bLen);
}
else if (bLen > aLen)
{
a = addZeros(a, bLen - aLen);
}
// Updated length
int len = Math.Max(aLen, bLen);
// To store the resultant XOR
String res = "";
for (int i = 0; i < len; i++)
{
if (a[i] == b[i])
res += "0";
else
res += "1";
}
return res;
}
// Driver code
public static void Main(String[] args)
{
String a = "11001", b = "111111";
Console.WriteLine(getXOR(a, b));
}
}
// This code is contributed by Rajput-Ji
<script>
// Javascript implementation of the approach
// Function to insert n 0s in the
// beginning of the given string
function addZeros(str, n)
{
for (let i = 0; i < n; i++)
{
str = "0" + str;
}
return str;
}
// Function to return the XOR
// of the given strings
function getXOR(a, b)
{
// Lengths of the given strings
let aLen = a.length;
let bLen = b.length;
// Make both the strings of equal lengths
// by inserting 0s in the beginning
if (aLen > bLen)
{
a = addZeros(b, aLen - bLen);
}
else if (bLen > aLen)
{
a = addZeros(a, bLen - aLen);
}
// Updated length
let len = Math.max(aLen, bLen);
// To store the resultant XOR
let res = "";
for (let i = 0; i < len; i++)
{
if (a[i] == b[i])
res += "0";
else
res += "1";
}
return res;
}
let a = "11001", b = "111111";
document.write(getXOR(a, b));
// This code is contributed by divyeshrabadiya07.
</script>
Time Complexity: O(len), len=Max(length a,length b)
Auxiliary Space: O(len)
Approach:
In this approach, we first determine the length of the larger string and iterate over it. For each index, we extract the corresponding bits from both input strings and perform a bitwise XOR operation on them. We then append the result to the output string. Finally, we return the output string.
Note that we use the expression (i < n) ? a[n - i - 1] - '0' : 0 to extract the bit at index i from the string a. This is because the bits in a are stored in reverse order compared to their position in the binary number. So we need to access the bits in reverse order by subtracting i from the length of the string and subtracting 1 (since the index starts from 0). We also subtract '0' from the bit to convert it from a character to an integer. Similarly, we extract the corresponding bit from b and perform the XOR operation.
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 XOR
// of the given strings
string getXOR(string a, string b)
{
string result = "";
int n = a.length();
int m = b.length();
int len = max(n, m);
for(int i = 0; i < len; i++) {
int x = (i < n) ? a[n - i - 1] - '0' : 0;
int y = (i < m) ? b[m - i - 1] - '0' : 0;
int z = x ^ y;
result = (char)(z + '0') + result;
}
return result;
}
// Driver code
int main()
{
string a = "11001", b = "111111";
cout << getXOR(a, b);
return 0;
}
public class GFG {
// Function to return the XOR of two binary strings
public static String getXOR(String a, String b) {
String result = "";
int n = a.length();
int m = b.length();
int len = Math.max(n, m);
for (int i = 0; i < len; i++) {
int x = (i < n) ? a.charAt(n - i - 1) - '0' : 0;
int y = (i < m) ? b.charAt(m - i - 1) - '0' : 0;
int z = x ^ y;
result = (char) (z + '0') + result;
}
return result;
}
public static void main(String[] args) {
String a = "11001";
String b = "111111";
System.out.println(getXOR(a, b));
}
}
# Function to return the XOR of the given binary strings
def get_xor(a, b):
result = "" # Initialize an empty string to store the XOR result
n = len(a)
m = len(b)
length = max(n, m)
for i in range(length): # Iterate through each bit position
x = int(a[n - i - 1]) if i < n else 0 # Get i-th bit of 'a' or 0 if it doesn't exist
y = int(b[m - i - 1]) if i < m else 0 # Get i-th bit of 'b' or 0 if it doesn't exist
z = x ^ y # Calculate XOR of x and y
result = str(z) + result # Prepend the XOR result to the 'result' string
return result
# Driver code
def main():
a = "11001"
b = "111111"
print(get_xor(a, b)) # Print the XOR of 'a' and 'b'
if __name__ == "__main__":
main()
using System;
class GFG {
// Function to return the XOR of the given binary
// strings
static string GetXOR(string a, string b)
{
string result = ""; // Initialize an empty string to
// store the XOR result
int n = a.Length; // Get the length of string 'a'
int m = b.Length; // Get the length of string 'b'
int len = Math.Max(n, m); // Find the maximum length
// between 'a' and 'b'
for (int i = 0; i < len; i++) {
// Extract the binary digits (0 or 1) from the
// right side of 'a' and 'b'
int x = (i < n) ? a[n - i - 1] - '0' : 0;
int y = (i < m) ? b[m - i - 1] - '0' : 0;
// Calculate the XOR of corresponding digits
int z = x ^ y;
// Convert the XOR result back to a character
// ('0' or '1') and prepend it to the result
// string
result = (char)(z + '0') + result;
}
return result; // Return the final XOR result as a
// binary string
}
// Driver code
static void Main()
{
string a = "11001"; // First binary string
string b = "111111"; // Second binary string
// Call the XOR function and print the result
Console.WriteLine(GetXOR(a, b));
}
}
// Function to return the XOR of the given strings
function getXOR(a, b) {
let result = "";
const n = a.length;
const m = b.length;
const len = Math.max(n, m);
for (let i = 0; i < len; i++) {
const x = i < n ? parseInt(a[n - i - 1]) : 0; // Get the current bit from string
// 'a' (right to left)
const y = i < m ? parseInt(b[m - i - 1]) : 0; // Get the current bit from string 'b'
// (right to left)
const z = x ^ y; // XOR operation between the two bits
result = z.toString() + result; // Add the result bit to the final string
}
return result;
}
// Driver code
const a = "11001";
const b = "111111";
console.log(getXOR(a, b)); // Output the XOR result
Time Complexity: O(max(n, m)), where n and m are the lengths of the input strings A and B, respectively.
Space Complexity: O(max(n, m)), as we need to create a new string to store the XOR of the input strings.
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