Find Maximum XOR value of a sub-array of size k

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Given an array of integers, the task is to find maximum XOR value of a subarray of size K.

Examples :

Input  : arr[] = {2, 5, 8 ,1 , 1 ,3} k = 3
Output : 15
Explanation : All subarrays of size k (=3) and
              their XOR values are:
 {2, 5, 8} => XOR value =  15
 {5, 8, 1} => XOR value =  12
 {8, 1, 1} => XOR value =  8
 {1, 1, 3} => XOR value =  3
Maximum of all XOR values = 15

Input  : arr[] = {1, 2, 4, 5, 6}
Output : 6

A simple solution is to consider all subarrays of size k one by one and compute XOR value. Finally return maximum of all XOR values. This solution takes O(n*k) time.

An efficient solution takes O(n) time. The idea is simple, we can find XOR value of current subarray of size k by removing first element of previous subarray and adding last element of current subarray. We can remove an element from current XOR by doing XOR of it again because of property of XOR that a ^ x ^ x = a.

Algorithm :

Let input array be 'arr[]' and size of array be 'n'

max_xor ;  // user to store maximum xor value
current_xor; //  user to store xor value of current subarray 
            // of size k 
 
// First compute xor value of first subarray of size k  
// (i goes from 0 to k)
corrent_xor = current_xor ^ arr[i] 

// Initialize maximum XOR
max_xor = current_xor 

Traversal rest array (i goes from k to n-1 )
 a).  remove first element of previous subarray 
      current_xor = current_xor ^ arr[i-k] 
 
 b).  add new element to subarray  
      current_xor = current_xor ^ arr[i]

 c). update max_xor = max(max_xor, current_xor)
 
return max_xor

Below is the implementation of above steps.

C++

// C++/C program to find maximum xor value of subarray of
// size k
#include<iostream>
using namespace std;
 
// Returns maximum XOR value of subarray of size k
int maximumXOR(int arr[] , int n , int k)
{
    // Compute XOR value of first subarray of size k
    int current_xor = 0 ;
    for (int i = 0 ; i < k ; i++)
        current_xor  = current_xor ^ arr[i];
 
    // Traverse rest of the array
    int max_xor = current_xor;
    for (int i = k ; i < n; i++)
    {
        // First remove previous subarray's first
        // element
        current_xor = current_xor ^ arr[i-k];
 
        // add new element
        current_xor = current_xor ^ arr[i];
 
        // Update maximum xor value
        max_xor = max(max_xor, current_xor);
    }
 
    return max_xor;
}
 
// Driver program
int main()
{
    int arr[] = {2, 5, 8 ,1 , 1 ,3} ;
    int k = 3;
    int n = sizeof(arr)/sizeof(arr[0]);
    cout << "Maximum XOR : " << maximumXOR(arr, n, k);
    return 0;
}

Java

// Java program to find maximum xor value of
// subarray of size k
import java.io.*;
 
class GFG {
 
    // Returns maximum XOR value of subarray of size k
    static int maximumXOR(int arr[] , int n , int k)
    {
         
        // Compute XOR value of first subarray of size k
        int current_xor = 0 ;
        for (int i = 0 ; i < k ; i++)
            current_xor = current_xor ^ arr[i];
     
        // Traverse rest of the array
        int max_xor = current_xor;
         
        for (int i = k ; i < n; i++)
        {
             
            // First remove previous subarray's first
            // element
            current_xor = current_xor ^ arr[i-k];
     
            // add new element
            current_xor = current_xor ^ arr[i];
     
            // Update maximum xor value
            max_xor = Math.max(max_xor, current_xor);
        }
     
        return max_xor;
    }
     
    // Driver program
    public static void main (String[] args)
    {
        int arr[] = {2, 5, 8 ,1 , 1 ,3} ;
        int k = 3;
        int n = arr.length;
        System.out.println( "Maximum XOR : "
                   + maximumXOR(arr, n, k));
    }
}
 
// This code is contributed by anuj_67.

Python 3

# Python3 program to find maximum 
# xor value of subarray of
# size
 
# Returns maximum XOR value 
# of subarray of size k
def maximumXOR(arr , n , k):
 
    # Compute XOR value of first 
    # subarray of size k
    current_xor = 0
    for i in range ( k):
        current_xor = current_xor ^ arr[i]
 
    # Traverse rest of the array
    max_xor = current_xor
    for i in range( k,n):
     
        # First remove previous subarray's first
        # element
        current_xor = current_xor ^ arr[i-k]
 
        # add new element
        current_xor = current_xor ^ arr[i]
 
        # Update maximum xor value
        max_xor = max(max_xor, current_xor)
     
 
    return max_xor
 
# Driver program
if __name__ =="__main__":
 
    arr = [2, 5, 8 ,1 , 1 ,3]
    k = 3
    n = len(arr)
    print ("Maximum XOR : "
          ,maximumXOR(arr, n, k))
 
# This code is contributed by
# ChitraNayal

C#

// C# program to find maximum
// xor value of subarray of 
// size k
using System;
class GFG {
 
    // Returns maximum XOR value
    // of subarray of size k
    static int maximumXOR(int []arr, 
                      int n, int k)
    {
         
        // Compute XOR value of first
        // subarray of size k
        int current_xor = 0 ;
        for (int i = 0; i < k; i++)
            current_xor = current_xor ^ arr[i];
     
        // Traverse rest of the array
        int max_xor = current_xor;
         
        for (int i = k ; i < n; i++)
        {
             
            // First remove previous
            // subarray's first
            // element
            current_xor = current_xor ^ arr[i-k];
     
            // add new element
            current_xor = current_xor ^ arr[i];
     
            // Update maximum xor value
            max_xor = Math.Max(max_xor, current_xor);
        }
     
        return max_xor;
    }
     
    // Driver Code
    public static void Main ()
    {
        int []arr = {2, 5, 8 ,1 , 1 ,3} ;
        int k = 3;
        int n = arr.Length;
        Console.WriteLine("Maximum XOR : "
                  + maximumXOR(arr, n, k));
    }
}
 
// This code is contributed by anuj_67.

PHP

<?php
// PHP program to find maximum 
// xor value of subarray of size k
 
// Returns maximum XOR value
// of subarray of size k
 
function maximumXOR($arr, $n, $k)
{
    // Compute XOR value of 
    // first subarray of size k
    $current_xor = 0 ;
    for ($i = 0 ; $i < $k ; $i++)
        $current_xor = $current_xor ^ 
                       $arr[$i];
 
    // Traverse rest of the array
    $max_xor = $current_xor;
    for ($i = $k ; $i < $n; $i++)
    {
        // First remove previous 
        // subarray's first element
        $current_xor = $current_xor ^ 
                       $arr[$i - $k];
 
        // add new element
        $current_xor = $current_xor ^ 
                       $arr[$i];
 
        // Update maximum xor value
        $max_xor = max($max_xor, 
                       $current_xor);
    }
 
    return $max_xor;
}
 
// Driver Code
$arr = array(2, 5, 8, 1, 1, 3) ;
$k = 3;
$n = sizeof($arr);
echo "Maximum XOR : ", 
      maximumXOR($arr, $n, $k);
 
// This code is contributed by m_kit
?>

Javascript

<script>
// Javascript program to find maximum xor value of
// subarray of size k
     
    // Returns maximum XOR value of subarray of size k    
    function maximumXOR(arr,n,k)
    {
        // Compute XOR value of first subarray of size k
        let current_xor = 0 ;
        for (let i = 0 ; i < k ; i++)
            current_xor = current_xor ^ arr[i];
       
        // Traverse rest of the array
        let max_xor = current_xor;
           
        for (let i = k ; i < n; i++)
        {
               
            // First remove previous subarray's first
            // element
            current_xor = current_xor ^ arr[i-k];
       
            // add new element
            current_xor = current_xor ^ arr[i];
       
            // Update maximum xor value
            max_xor = Math.max(max_xor, current_xor);
        }
       
        return max_xor;
    }
     
    // Driver program
    let arr=[2, 5, 8 ,1 , 1 ,3];
    let k = 3;
    let  n = arr.length;
    document.write( "Maximum XOR : "
                   + maximumXOR(arr, n, k));
     
     
    // This code is contributed by rag2127
</script>

Output

Maximum XOR : 15

Time Complexity : O(n)

Auxiliary Space: O(1)

This article is contributed by Nishant_Singh(Pintu). If you like zambiatek and would like to contribute, you can also write an article using write.zambiatek.com or mail your article to review-team@zambiatek.com. See your article appearing on the zambiatek main page and help other Geeks.

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