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# Bitwise XOR of a Binary array

• Difficulty Level : Easy
• Last Updated : 21 May, 2021

Given a binary array arr[], the task is to calculate the bitwise XOR of all the elements in this array and print it.

Examples:

Input: arr[] = {“100”, “1001”, “0011”}
Output: 1110
0100 XOR 1001 XOR 0011 = 1110

Input: arr[] = {“10”, “11”, “1000001”}
Output: 1000000

Approach:

• Step 1: First find the maximum-sized binary string.
• Step 2: Make all the binary strings in an array to the size of the maximum sized string, by adding 0s at the Most Significant Bit
• Step 3: Now find the resultant string by performing bitwise XOR on all the binary strings in the array.

For Examples:

1. Let the binary array be {“100”, “001”, and “1111”}.
2. Here the maximum sized binary string is 4.
3. Make all the binary strings in the array of size 4, by adding 0s at the MSB. Now the binary array becomes {“0100”, “0001” and “1111”}
4. Performing bitwise XOR on all the binary strings in the array
`“0100” XOR “0001” XOR “1111” = “1110”`

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach` `#include ``using` `namespace` `std;` `// Function to return the bitwise XOR``// of all the binary strings``void` `strBitwiseXOR(string* arr, ``int` `n)``{``    ``string result;` `    ``int` `max_len = INT_MIN;` `    ``// Get max size and reverse each string``    ``// Since we have to perform XOR operation``    ``// on bits from right to left``    ``// Reversing the string will make it easier``    ``// to perform operation from left to right``    ``for` `(``int` `i = 0; i < n; i++) {``        ``max_len = max(max_len,``                      ``(``int``)arr[i].size());``        ``reverse(arr[i].begin(),``                ``arr[i].end());``    ``}` `    ``for` `(``int` `i = 0; i < n; i++) {` `        ``// Add 0s to the end``        ``// of strings if needed``        ``string s;``        ``for` `(``int` `j = 0;``             ``j < max_len - arr[i].size();``             ``j++)``            ``s += ``'0'``;` `        ``arr[i] = arr[i] + s;``    ``}` `    ``// Perform XOR operation on each bit``    ``for` `(``int` `i = 0; i < max_len; i++) {``        ``int` `pres_bit = 0;` `        ``for` `(``int` `j = 0; j < n; j++)``            ``pres_bit = pres_bit ^ (arr[j][i] - ``'0'``);` `        ``result += (pres_bit + ``'0'``);``    ``}` `    ``// Reverse the resultant string``    ``// to get the final string``    ``reverse(result.begin(), result.end());` `    ``// Return the final string``    ``cout << result;``}` `// Driver code``int` `main()``{``    ``string arr[] = { ``"1000"``, ``"10001"``, ``"0011"` `};``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr);` `    ``strBitwiseXOR(arr, n);``    ``return` `0;``}`

## Java

 `// Java implementation of the approach``import` `java.io.*;``import` `java.util.*;` `class` `GFG{` `// Function to return the``// reverse string``static` `String reverse(String str)``{``    ``String rev = ``""``;``    ``for``(``int` `i = str.length() - ``1``; i >= ``0``; i--)``        ``rev = rev + str.charAt(i);` `    ``return` `rev;``}` `// Function to return the bitwise XOR``// of all the binary strings``static` `String strBitwiseXOR(String[] arr, ``int` `n)``{``    ``String result = ``""``;` `    ``int` `max_len = Integer.MIN_VALUE;` `    ``// Get max size and reverse each string``    ``// Since we have to perform XOR operation``    ``// on bits from right to left``    ``// Reversing the string will make it easier``    ``// to perform operation from left to right``    ``for``(``int` `i = ``0``; i < n; i++)``    ``{``        ``max_len = Math.max(max_len,``                  ``(``int``)arr[i].length());``        ``arr[i] = reverse(arr[i]);``    ``}` `    ``for``(``int` `i = ``0``; i < n; i++)``    ``{``        ` `        ``// Add 0s to the end``        ``// of strings if needed``        ``String s = ``""``;``        ``for``(``int` `j = ``0``;``                ``j < max_len - arr[i].length();``                ``j++)``            ``s += ``'0'``;` `        ``arr[i] = arr[i] + s;``    ``}` `    ``// Perform XOR operation on each bit``    ``for``(``int` `i = ``0``; i < max_len; i++)``    ``{``        ``int` `pres_bit = ``0``;` `        ``for``(``int` `j = ``0``; j < n; j++)``            ``pres_bit = pres_bit ^``                      ``(arr[j].charAt(i) - ``'0'``);` `        ``result += (``char``)(pres_bit + ``'0'``);``    ``}` `    ``// Reverse the resultant string``    ``// to get the final string``    ``result = reverse(result);` `    ``// Return the final string``    ``return` `result;``}` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``String[] arr = { ``"1000"``, ``"10001"``, ``"0011"` `};``    ``int` `n = arr.length;` `    ``System.out.print(strBitwiseXOR(arr, n));``}``}` `// This code is contributed by akhilsaini`

## Python3

 `# Function to return the bitwise XOR``# of all the binary strings``import` `sys``def` `strBitwiseXOR(arr, n):``    ``result ``=` `""``    ``max_len ``=` `-``1`` ` `    ``# Get max size and reverse each string``    ``# Since we have to perform XOR operation``    ``# on bits from right to left``    ``# Reversing the string will make it easier``    ``# to perform operation from left to right``    ``for` `i ``in` `range``(n):``        ``max_len ``=` `max``(max_len, ``len``(arr[i]))``        ``arr[i] ``=` `arr[i][::``-``1``]`` ` `    ``for` `i ``in` `range``(n):``        ``# Add 0s to the end``        ``# of strings if needed``        ``s ``=` `""``        ``# t = max_len - len(arr[i])``        ``for` `j ``in` `range``(max_len ``-` `len``(arr[i])):``            ``s ``+``=` `"0"`` ` `        ``arr[i] ``=` `arr[i] ``+` `s`` ` `    ``# Perform XOR operation on each bit``    ``for` `i ``in` `range``(max_len):``        ``pres_bit ``=` `0`` ` `        ``for` `j ``in` `range``(n):``            ``pres_bit ``=` `pres_bit ^ (``ord``(arr[j][i]) ``-` `ord``(``'0'``))`` ` `        ``result ``+``=` `chr``((pres_bit) ``+` `ord``(``'0'``))`` ` `    ``# Reverse the resultant string``    ``# to get the final string``    ``result ``=` `result[::``-``1``]`` ` `    ``# Return the final string``    ``print``(result)``  ` `# Driver code``if``(__name__ ``=``=` `"__main__"``):``    ``arr ``=` `[``"1000"``, ``"10001"``, ``"0011"``]``    ``n ``=` `len``(arr)``    ``strBitwiseXOR(arr, n)` `# This code is contributed by skylags`

## C#

 `// C# implementation of the approach``using` `System;` `class` `GFG{` `// Function to return the``// reverse string``static` `string` `reverse(``string` `str)``{``    ``string` `rev = ``""``;``    ``for``(``int` `i = str.Length - 1; i >= 0; i--)``        ``rev = rev + str[i];` `    ``return` `rev;``}` `// Function to return the bitwise XOR``// of all the binary strings``static` `string` `strBitwiseXOR(``string``[] arr, ``int` `n)``{``    ``string` `result = ``""``;` `    ``int` `max_len = ``int``.MinValue;` `    ``// Get max size and reverse each string``    ``// Since we have to perform XOR operation``    ``// on bits from right to left``    ``// Reversing the string will make it easier``    ``// to perform operation from left to right``    ``for``(``int` `i = 0; i < n; i++)``    ``{``        ``max_len = Math.Max(max_len,``                          ``(``int``)arr[i].Length);``        ``arr[i] = reverse(arr[i]);``    ``}` `    ``for``(``int` `i = 0; i < n; i++)``    ``{``        ` `        ``// Add 0s to the end``        ``// of strings if needed``        ``string` `s = ``""``;``        ``for``(``int` `j = 0;``                ``j < max_len - arr[i].Length;``                ``j++)``            ``s += ``'0'``;` `        ``arr[i] = arr[i] + s;``    ``}` `    ``// Perform XOR operation on each bit``    ``for``(``int` `i = 0; i < max_len; i++)``    ``{``        ``int` `pres_bit = 0;` `        ``for``(``int` `j = 0; j < n; j++)``            ``pres_bit = pres_bit ^``                       ``(arr[j][i] - ``'0'``);` `        ``result += (``char``)(pres_bit + ``'0'``);``    ``}``    ` `    ``// Reverse the resultant string``    ``// to get the final string``    ``result = reverse(result);` `    ``// Return the final string``    ``return` `result;``}` `// Driver code``public` `static` `void` `Main()``{``    ``string``[] arr = { ``"1000"``, ``"10001"``, ``"0011"` `};``    ``int` `n = arr.Length;` `    ``Console.Write(strBitwiseXOR(arr, n));``}``}` `// This code is contributed by akhilsaini`

## Javascript

 ``
Output:
`11010`

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