Count subarrays with sum equal to its XOR value

Given an array arr[] containing N elements, the task is to count the number of sub-arrays whose XOR of all the elements is equal to the sum of all the elements in the subarray.

Examples:

Input: arr[] = {2, 5, 4, 6}
Output: 5
Explanation:
All the subarrays {{2}, {5}, {4}, {6}} satisfies the above condition since the XOR of the subarrays is same as the sum. Apart from these, the subarray {2, 5} also satisfies the condition:
(2 xor 5) = 7 = (2 + 5)

Input: arr[] = {1, 2, 3, 4, 5}
Output: 7

Naive Approach: The naive approach for this problem is to consider all the sub-arrays and for every subarray, check if the XOR is equal to the sum.



Time Complexity: O(N2)

Efficient Approach: The idea is to use the concept of sliding window. First, we calculate the window for which the above condition is satisfied and then we slide through every element till N. The following steps can be followed to compute the answer:

  • Maintain two pointers left and right initially assigned to zero.
  • Calculate the window using right pointer where the condition A xor B = A + B is satisfied.
  • Count of the sub-arrays will be right – left.
  • Iterate through every element and remove the previous element.

Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ program to count the number
// of subarrays such that Xor of
// all the elements of that subarray
// is equal to sum of the elements
  
#include <bits/stdc++.h>
#define ll long long int
using namespace std;
  
// Function to count the number
// of subarrays such that Xor of
// all the elements of that subarray
// is equal to sum of the elements
ll operation(int arr[], int N)
{
    // Maintain two pointers
    // left and right
    ll right = 0, ans = 0,
       num = 0;
  
    // Iterating through the array
    for (ll left = 0; left < N; left++) {
  
        // Calculate the window
        // where the above condition
        // is satisfied
        while (right < N
               && num + arr[right]
                      == (num ^ arr[right])) {
            num += arr[right];
            right++;
        }
  
        // Count will be (right-left)
        ans += right - left;
        if (left == right)
            right++;
  
        // Remove the previous element
        // as it is already included
        else
            num -= arr[left];
    }
  
    return ans;
}
  
// Driver code
int main()
{
    int arr[] = { 1, 2, 3, 4, 5 };
    int N = sizeof(arr) / sizeof(arr[0]);
  
    cout << operation(arr, N);
}

chevron_right






                        filter_none









Java














                                    filter_none




                                    edit

                                    close




                                    play_arrow




                                    link

                                    brightness_4

                                    code
                                




















// Java program to count the number 


// of subarrays such that Xor of all  


// the elements of that subarray is  


// equal to sum of the elements 


import java.io.*; 


  


class GFG{ 


      


// Function to count the number 


// of subarrays such that Xor of 


// all the elements of that subarray 


// is equal to sum of the elements 


static long operation(int arr[], int N) 


{ 


      


    // Maintain two pointers 


    // left and right 


    int right = 0; 


    int    num = 0; 


    long ans = 0; 


  


    // Iterating through the array 


    for(int left = 0; left < N; left++)  


    { 


         


       // Calculate the window 


       // where the above condition 


       // is satisfied 


       while (right < N && num + arr[right] ==  


                          (num ^ arr[right])) 


       { 


           num += arr[right]; 


           right++; 


       } 


         


       // Count will be (right-left) 


       ans += right - left; 


       if (left == right) 


           right++; 


         


       // Remove the previous element 


       // as it is already included 


       else


           num -= arr[left]; 


    } 


    return ans; 


} 


  


// Driver code 


public static void main(String[] args) 


{ 


    int arr[] = { 1, 2, 3, 4, 5 }; 


    int N = arr.length; 


  


    System.out.println(operation(arr, N)); 


} 


} 


  


// This code is contributed by offbeat 



















                        chevron_right







                        filter_none









Python3














                                    filter_none




                                    edit

                                    close




                                    play_arrow




                                    link

                                    brightness_4

                                    code
                                




















# Python3 program to count the number 


# of subarrays such that Xor of 


# all the elements of that subarray 


# is equal to sum of the elements 


  


# Function to count the number 


# of subarrays such that Xor of 


# all the elements of that subarray 


# is equal to sum of the elements 


def operation(arr, N): 


  


    # Maintain two pointers 


    # left and right 


    right = 0; ans = 0; 


    num = 0; 


  


    # Iterating through the array 


    for left in range(0, N): 


  


        # Calculate the window 


        # where the above condition 


        # is satisfied 


        while (right < N and 


               num + arr[right] == 


              (num ^ arr[right])): 


            num += arr[right]; 


            right += 1; 


  


        # Count will be (right-left) 


        ans += right - left; 


        if (left == right): 


            right += 1; 


  


        # Remove the previous element 


        # as it is already included 


        else: 


            num -= arr[left]; 


  


    return ans; 


  


# Driver code 


arr = [1, 2, 3, 4, 5]; 


N = len(arr) 


print(operation(arr, N)); 


  


# This code is contributed by Nidhi_biet 



















                        chevron_right







                        filter_none









C#














                                    filter_none




                                    edit

                                    close




                                    play_arrow




                                    link

                                    brightness_4

                                    code
                                




















// C# program to count the number 


// of subarrays such that Xor of all  


// the elements of that subarray is  


// equal to sum of the elements 


using System; 


class GFG{ 


      


// Function to count the number 


// of subarrays such that Xor of 


// all the elements of that subarray 


// is equal to sum of the elements 


static long operation(int []arr, int N) 


{ 


      


    // Maintain two pointers 


    // left and right 


    int right = 0; 


    int num = 0; 


    long ans = 0; 


  


    // Iterating through the array 


    for(int left = 0; left < N; left++)  


    { 


          


        // Calculate the window 


        // where the above condition 


        // is satisfied 


        while (right < N &&  


               num + arr[right] ==  


              (num ^ arr[right])) 


        { 


            num += arr[right]; 


            right++; 


        } 


              


        // Count will be (right-left) 


        ans += right - left; 


        if (left == right) 


            right++; 


              


        // Remove the previous element 


        // as it is already included 


        else


            num -= arr[left]; 


    } 


    return ans; 


} 


  


// Driver code 


public static void Main(String[] args) 


{ 


    int []arr = { 1, 2, 3, 4, 5 }; 


    int N = arr.Length; 


  


    Console.WriteLine(operation(arr, N)); 


} 


} 


  


// This code is contributed by 29AjayKumar 



















                        chevron_right







                        filter_none











Output:


7





Time Complexity: O(N), where N is the length of the array.


competitive-programming-img





    

        My Personal Notes
        arrow_drop_up
    

    

        
        

            
            
        

    


Recommended Posts:
godaraji47
Check out this Author's contributed articles.
If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please Improve this article if you find anything incorrect by clicking on  the "Improve Article" button below.