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 stringvoid addZeros(string& str, int n){ for (int i = 0; i < n; i++) { str = "0" + str; }}// Function to return the XOR// of the given stringsstring 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 codeint main(){ string a = "11001", b = "111111"; cout << getXOR(a, b); return 0;} |
Java
// 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
# Python3 implementation of the approach# Function to insert n 0s in the# beginning of the given stringdef addZeros(strr, n): for i in range(n): strr = "0" + strr return strr# Function to return the XOR# of the given stringsdef 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 codea = "11001"b = "111111"print(getXOR(a, b))# This code is contributed by Mohit Kumar |
C#
// 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 |
Javascript
<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> |
Output
100110
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 stringsstring 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 codeint main(){ string a = "11001", b = "111111"; cout << getXOR(a, b); return 0;} |
Java
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)); }} |
Python3
# Function to return the XOR of the given binary stringsdef 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 codedef main(): a = "11001" b = "111111" print(get_xor(a, b)) # Print the XOR of 'a' and 'b'if __name__ == "__main__": main() |
C#
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)); }} |
Output
100110
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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