Sum of XOR of all sub-arrays of length K

Given an array of length n ( n > k), we have to find the sum of xor of all the elements of the sub-arrays which are of length k.
Examples:

Input : arr[]={1, 2, 3, 4}, k=2 
Output :Sum= 11 
Sum = 1^2 + 2^3 + 3^4 = 3 + 1 + 7 =11
Input :arr[]={1, 2, 3, 4}, k=3 
Output :Sum= 5 
Sum = 1^2^3 + 2^3^4 = 0 + 5 =5 

Naive Solution: The idea is to traverse all the subarrays of length k and find the xor of all the elements of the subarray and sum them up to find the sum of XOR of all K length sub-array of an array. 
Time Complexity:  O(N²)
Auxiliary Space: O(1)

Efficient Solution: The efficient solution is to traverse the array and find all the subarray of length k, i.e. ( 0 to k-1), (1 to k), (2 to k+1), …., (n-k+1 to n).
We will find and store the xor of elements from 0 to i (in an array x[]) by forming a pre-xor array.
Now, xor of sub array from l to r is equal to x[l-1] ^ x[r] because x[r] will give the xor of all elements till r and x[l-1] will give the xor of all elements till l-1. When we will take xor of these two values the elements till 0 to l-1 will be repeated. As a^a = 0, the repeated values would contribute zero to the net value and we get the value of xor sub array from l to r.
Below is the implementation of the above approach: 

C++

// C++ implementation of above approach
#include <iostream>
using namespace std;

// Sum of XOR of all K length
// sub-array of an array
int FindXorSum(int arr[], int k, int n)
{
    // If the length of the array is less than k
    if (n < k)
        return 0;

    // Array that will store xor values of
    // subarray from 1 to i
    int x[n] = { 0 };
    int result = 0;

    // Traverse through the array
    for (int i = 0; i < n; i++) {

        // If i is greater than zero, store
        // xor of all the elements from 0 to i
        if (i > 0)
            x[i] = x[i - 1] ^ arr[i];

        // If it is the first element
        else
            x[i] = arr[i];

        // If i is greater than k
        if (i >= k - 1) {
            int sum = 0;

            // Xor of values from 0 to i
            sum = x[i];

            // Now to find subarray of length k
            // that ends at i, xor sum with x[i-k]
            if (i - k > -1)
                sum ^= x[i - k];

            // Add the xor of elements from i-k+1 to i
            result += sum;
        }
    }

    // Return the resultant sum;
    return result;
}

// Driver code
int main()
{
    int arr[] = { 1, 2, 3, 4 };

    int n = 4, k = 2;

    cout << FindXorSum(arr, k, n) << endl;

    return 0;
}

Java

// Java implementation of the approach
class GFG
{

// Sum of XOR of all K length
// sub-array of an array
static int FindXorSum(int arr[], int k, int n)
{
    // If the length of the array is less than k
    if (n < k)
        return 0;

    // Array that will store xor values of
    // subarray from 1 to i
    int []x = new int[n];
    int result = 0;

    // Traverse through the array
    for (int i = 0; i < n; i++)
    {

        // If i is greater than zero, store
        // xor of all the elements from 0 to i
        if (i > 0)
            x[i] = x[i - 1] ^ arr[i];

        // If it is the first element
        else
            x[i] = arr[i];

        // If i is greater than k
        if (i >= k - 1)
        {
            int sum = 0;

            // Xor of values from 0 to i
            sum = x[i];

            // Now to find subarray of length k
            // that ends at i, xor sum with x[i-k]
            if (i - k > -1)
                sum ^= x[i - k];

            // Add the xor of elements from i-k+1 to i
            result += sum;
        }
    }

    // Return the resultant sum;
    return result;
}

// Driver code
public static void main(String[] args)
{
    int arr[] = { 1, 2, 3, 4 };

    int n = 4, k = 2;

    System.out.println(FindXorSum(arr, k, n));
}
}

// This code contributed by Rajput-Ji

Python3

# Python implementation of above approach

# Sum of XOR of all K length
# sub-array of an array
def FindXorSum(arr, k, n):

    # If the length of the array is less than k
    if (n < k):
        return 0;

    # Array that will store xor values of
    # subarray from 1 to i
    x = [0]*n;
    result = 0;

    # Traverse through the array
    for i in range(n):

        # If i is greater than zero, store
        # xor of all the elements from 0 to i
        if (i > 0):
            x[i] = x[i - 1] ^ arr[i];

        # If it is the first element
        else:
            x[i] = arr[i];

        # If i is greater than k
        if (i >= k - 1):
            sum = 0;

            # Xor of values from 0 to i
            sum = x[i];

            # Now to find subarray of length k
            # that ends at i, xor sum with x[i-k]
            if (i - k > -1):
                sum ^= x[i - k];

            # Add the xor of elements from i-k+1 to i
            result += sum;

    # Return the resultant sum;
    return result;

# Driver code
arr = [ 1, 2, 3, 4 ];

n = 4; k = 2;

print(FindXorSum(arr, k, n));

# This code has been contributed by 29AjayKumar

C#

// C# implementation of the above approach
using System;

class GFG
{

    // Sum of XOR of all K length
    // sub-array of an array
    static int FindXorSum(int []arr, int k, int n)
    {
        // If the length of the array is less than k
        if (n < k)
            return 0;

        // Array that will store xor values of
        // subarray from 1 to i
        int []x = new int[n];
        int result = 0;

        // Traverse through the array
        for (int i = 0; i < n; i++)
        {

            // If i is greater than zero, store
            // xor of all the elements from 0 to i
            if (i > 0)
                x[i] = x[i - 1] ^ arr[i];

            // If it is the first element
            else
                x[i] = arr[i];

            // If i is greater than k
            if (i >= k - 1)
            {
                int sum = 0;

                // Xor of values from 0 to i
                sum = x[i];

                // Now to find subarray of length k
                // that ends at i, xor sum with x[i-k]
                if (i - k > -1)
                    sum ^= x[i - k];

                // Add the xor of elements from i-k+1 to i
                result += sum;
            }
        }

        // Return the resultant sum;
        return result;
    }

    // Driver code
    public static void Main()
    {
        int []arr = { 1, 2, 3, 4 };

        int n = 4, k = 2;

        Console.WriteLine(FindXorSum(arr, k, n));
    }
}

// This code is contributed by AnkitRai01

Javascript

<script>

// Javascript implementation of above approach

// Sum of XOR of all K length
// sub-array of an array
function FindXorSum(arr, k, n)
{
    // If the length of the array is less than k
    if (n < k)
        return 0;

    // Array that will store xor values of
    // subarray from 1 to i
    let x = new Array(n).fill(0);
    let result = 0;

    // Traverse through the array
    for (let i = 0; i < n; i++) {

        // If i is greater than zero, store
        // xor of all the elements from 0 to i
        if (i > 0)
            x[i] = x[i - 1] ^ arr[i];

        // If it is the first element
        else
            x[i] = arr[i];

        // If i is greater than k
        if (i >= k - 1) {
            let sum = 0;

            // Xor of values from 0 to i
            sum = x[i];

            // Now to find subarray of length k
            // that ends at i, xor sum with x[i-k]
            if (i - k > -1)
                sum ^= x[i - k];

            // Add the xor of elements from i-k+1 to i
            result += sum;
        }
    }

    // Return the resultant sum;
    return result;
}

// Driver code
    let arr = [ 1, 2, 3, 4 ];

    let n = 4, k = 2;

    document.write(FindXorSum(arr, k, n));

</script>

Output:

11

Time Complexity: O(n)
Auxiliary Space: O(n)