Bitwise XOR of a Binary array - GeeksforGeeks

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: 
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
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++
Java
Python3
C#
Javascript
// C++ implementation of the approach #include <bits/stdc++.h>using namespace std; // Function to return the bitwise XOR// of all the binary stringsvoid 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 codeint main(){    string arr[] = { "1000", "10001", "0011" };    int n = sizeof(arr) / sizeof(arr[0]);     strBitwiseXOR(arr, n);    return 0;}
// Java implementation of the approachimport java.io.*;import java.util.*; class GFG{ // Function to return the// reverse stringstatic 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 stringsstatic 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 codepublic 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
# Function to return the bitwise XOR# of all the binary stringsimport sysdef 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 codeif(__name__ == "__main__"):    arr = ["1000", "10001", "0011"]    n = len(arr)    strBitwiseXOR(arr, n) # This code is contributed by skylags
// C# implementation of the approachusing System; class GFG{ // Function to return the// reverse stringstatic 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 stringsstatic 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 codepublic static void Main(){    string[] arr = { "1000", "10001", "0011" };    int n = arr.Length;     Console.Write(strBitwiseXOR(arr, n));}} // This code is contributed by akhilsaini
<script> // Javascript implementation of the approach // Function to return the bitwise XOR// of all the binary stringsfunction 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 codevar arr = ["1000", "10001", "0011"];var n = arr.length;strBitwiseXOR(arr, n); </script>
11010
 
skylags
akhilsaini
noob2000
binary-string
Bitwise-XOR
Arrays
Bit Magic
Strings
Arrays
Strings
Bit Magic
Writing code in comment?
Please use ide.geeksforgeeks.org,
generate link and share the link here.
Stack Data Structure (Introduction and Program)
Maximum and minimum of an array using minimum number of comparisons
Top 50 Array Coding Problems for Interviews
Multidimensional Arrays in Java
Introduction to Arrays
Bitwise Operators in C/C++
Left Shift and Right Shift Operators in C/C++
Travelling Salesman Problem | Set 1 (Naive and Dynamic Programming)
Count set bits in an integer
Cyclic Redundancy Check and Modulo-2 Division