# 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: 100110

Input: A = “11111”, B = “0”
Output: 11111

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ` `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

 `// 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 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 `

## 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 `

Output:

```100110
``` My Personal Notes arrow_drop_up Check out this Author's contributed articles.

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