XOR two binary strings of unequal lengths
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: 100110Input: 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; } |
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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 |
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Python3
# Python3 implementation of the approach # Function to insert n 0s in the # beginning of the given strring def addZeros(strr, n): for i in range(n): strr = "0" + strr return strr # Function to return the XOR # of the given strrings def getXOR(a, b): # Lengths of the given strrings aLen = len(a) bLen = len(b) # Make both the strrings 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 |
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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 |
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Output:
100110
