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)

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++

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// 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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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 

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Output:

11

Time Complexity : O(N)



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