Bitwise XOR of a Binary array

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 <bits/stdc++.h>
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[0]);

    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

<script>

// Javascript implementation of the approach

// Function to return the bitwise XOR
// of all the binary strings
function strBitwiseXOR(arr, n)
{
    var result = "";

    var max_len = -1000000000;

    // 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 (var i = 0; i < n; i++) {
        max_len = Math.max(max_len,
                       arr[i].length);
        arr[i] = arr[i].split('').reverse().join('');
    }

    for (var i = 0; i < n; i++) {

        // Add 0s to the end
        // of strings if needed
        var s;
        for (var j = 0;
             j < max_len - arr[i].length;
             j++)
            s += '0';

        arr[i] = arr[i] + s;
    }

    // Perform XOR operation on each bit
    for (var i = 0; i < max_len; i++) {
        var pres_bit = 0;

        for (var j = 0; j < n; j++)
            pres_bit = pres_bit ^ (arr[j][i] - '0'.charCodeAt(0));

        result += (pres_bit + '0'.charCodeAt(0));
    }

    // Reverse the resultant string
    // to get the final string
    result = result.split('').reverse().join('');

    // Return the final string
    document.write( result);
}

// Driver code
var arr = ["1000", "10001", "0011"];
var n = arr.length;
strBitwiseXOR(arr, n);

</script>

Output:

11010

Time complexity: O(n*m) where n is the number of strings in the array, and m is the maximum length of any string in the array.

Space Complexity: O(n*m)