Bitwise XOR of a Binary array

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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:

Let the binary array be {“100”, “001”, and “1111”}.
Here the maximum sized binary string is 4.
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”}
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:

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)

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