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Add an element in Array to make the bitwise XOR as K

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  • Last Updated : 02 May, 2022

Given an array arr[] containing N positive integers, the task is to add an integer such that the bitwise Xor of the new array becomes K.

Examples:

Input: arr[] = {1, 4, 5, 6}, K = 4
Output: 2
Explanation: Bit-wise XOR of the array is 6. 
And bit-wise XOR of 6 and 2 is 4.

Input: arr[] = {2, 7, 9, 1}, K = 5
Output: 8

 

Approach: The solution to the problem is based on the following idea of bitwise Xor:

If for two numbers X and Y, the bitwise Xor of X and Y is Z then the bitwise Xor of X and Z is Y.

Follow the steps to solve the problem:

  • Let the bitwise XOR of the array elements be X
  • Say the required value to be added is Y such that X Xor Y = K.
  • From the above observation, it is clear that the value to be added (Y) is the same as X Xor K.

Below is the implementation of the above approach:

C++




// C++ code to implement the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the required value
int find_K(int K, vector<int>& arr)
{
    int XOR = 0;
 
    // Find XOR of all the array elements
    for (int i = 0; i < arr.size(); i++)
        XOR ^= arr[i];
 
    // K = XOR^N, where N is size of array
    return XOR ^ K;
}
 
// Drivers code
int main()
{
    int K = 4;
    vector<int> arr = { 1, 4, 5, 6 };
 
    // Function call
    cout << find_K(K, arr);
    return 0;
}

Java




// Java code to implement the above approach
import java.io.*;
 
class GFG {
    // Function to find the required value
    public static int find_K(int K, int arr[])
    {
        int XOR = 0;
 
        // Find XOR of all the array elements
        for (int i = 0; i < arr.length; i++)
            XOR ^= arr[i];
 
        // K = XOR^N, where N is size of array
        return XOR ^ K;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int K = 4;
        int arr[] = { 1, 4, 5, 6 };
 
        // Function call
        System.out.print(find_K(K, arr));
    }
}
 
// This code is contributed by Rohit Pradhan

Python3




# Python code to implement the above approach
 
# Function to find the required value
def find_K(K, arr):
    XOR = 0
 
    # Find XOR of all the array elements
    for i in range(len(arr)):
        XOR ^= arr[i]
 
    # K = XOR^N, where N is size of array
    return XOR ^ K
 
# Drivers code
K = 4
arr = [ 1, 4, 5, 6 ]
 
# Function call
print(find_K(K, arr))
 
# This code is contributed by shinjanpatra

C#




// C# code to implement the above approach
using System;
 
class GFG
{
 
  // Function to find the required value
  static int find_K(int K, int[] arr)
  {
    int XOR = 0;
 
    // Find XOR of all the array elements
    for (int i = 0; i < arr.Length; i++)
      XOR ^= arr[i];
 
    // K = XOR^N, where N is size of array
    return XOR ^ K;
  }
 
  // Driver Code
  public static int Main()
  {
    int K = 4;
    int[] arr = new int[] { 1, 4, 5, 6 };
 
    // Function call
    Console.Write(find_K(K, arr));
    return 0;
  }
}
 
// This code is contributed by Taranpreet

Javascript




<script>
    // JavaScript code to implement the above approach
 
    // Function to find the required value
    const find_K = (K, arr) => {
        let XOR = 0;
 
        // Find XOR of all the array elements
        for (let i = 0; i < arr.length; i++)
            XOR = (XOR ^ arr[i]);
 
        // K = XOR^N, where N is size of array
        return (XOR ^ K);
    }
 
    // Drivers code
 
    let K = 4;
    let arr = [1, 4, 5, 6];
 
    // Function call
    document.write(find_K(K, arr));
 
// This code is contributed by rakeshsahni
 
</script>

Output

2

Time Complexity: O(N)
Auxiliary Space: O(1)


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