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Count subarrays with sum equal to its XOR value

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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:
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:

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. 

  • Generate all the subarray
  • Calculate the subarray sum and xor to all its elements
  • Check if the subarray sum is equal to xor of all its elements then increment the count
  • Finally, return count.

Below is the implementation of the above approach:

C++




// 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>
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
int operation(int arr[], int n)
{
    int count = 0;
    for(int i = 0; i < n ; i++){
        int sum = 0;
        int xorr = 0;
         
        for(int j = i; j < n; j++){
            sum += arr[j];
            xorr ^= arr[j];
 
            if(sum == xorr) count++;
        }
    }
    return count;
}
 
// Driver code
int main()
{
    int arr[] = { 1, 2, 3, 4, 5 };
    int N = sizeof(arr) / sizeof(arr[0]);
 
    cout << operation(arr, N);
}


Java




import java.util.*;
 
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
    public static int operation(int[] arr, int n)
    {
        int count = 0;
        for (int i = 0; i < n; i++) {
            int sum = 0;
            int xorr = 0;
 
            for (int j = i; j < n; j++) {
                sum += arr[j];
                xorr ^= arr[j];
 
                if (sum == xorr)
                    count++;
            }
        }
        return count;
    }
 
    public static void main(String[] args)
    {
        int[] arr = { 1, 2, 3, 4, 5 };
        int N = arr.length;
 
        System.out.println(operation(arr, N));
    }
}


Python3




# Python 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):
    count = 0;
    for i in range(0,n):
        sum = 0;
        xorr = 0;
         
        for j in range(i,n):
            sum += arr[j];
            xorr ^= arr[j];
 
            if(sum == xorr):
                count+=1;
         
    return count;
 
# Driver code
arr = [ 1, 2, 3, 4, 5 ];
N = len(arr);
 
print(operation(arr, N));


C#




using System;
using System.Linq;
 
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
    public static int operation(int[] arr, int n)
    {
        int count = 0;
        for (int i = 0; i < n; i++) {
            int sum = 0;
            int xorr = 0;
 
            for (int j = i; j < n; j++) {
                sum += arr[j];
                xorr ^= arr[j];
 
                if (sum == xorr)
                    count++;
            }
        }
        return count;
    }
 
    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 divya_p123.


Javascript




// Javascript 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
function operation( arr, n)
{
    let count = 0;
    for(let i = 0; i < n ; i++){
        let sum = 0;
        let xorr = 0;
         
        for(let j = i; j < n; j++){
            sum += arr[j];
            xorr ^= arr[j];
 
            if(sum == xorr)
                count++;
        }
    }
    return count;
}
 
// Driver code
    let arr = [1, 2, 3, 4, 5 ];
    let N = arr.length;
 
    document.write(operation(arr, N));


Output

7

Time Complexity: O(N2)
Auxiliary Space: O(1)

Efficient Approach: The idea is to use the concept of sliding window.

Below implementation uses two pointers, left and right, to define the current subarray. The num variable stores the XOR of all elements in the current subarray, and it is also used to keep track of the sum of the elements in the subarray.

The outer loop iterates left from 0 to N-1, while the inner loop moves right to the right until the condition num + arr[right] == (num ^ arr[right]) is no longer satisfied. The right – left is then added to ans to count the number of subarrays satisfying the condition.

If left == right, right is incremented. Otherwise, the element at index left is removed from num. The process is repeated for the next value of left.

Finally, the function returns ans, which is the number of subarrays satisfying the condition.

Below is the implementation of the above approach:

C++




// 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);
}


Java




// 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


Python3




# 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


C#




// 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


Javascript




<script>
 
// Javascript 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
function operation(arr, N)
{
    // Maintain two pointers
    // left and right
    let right = 0, ans = 0,
       num = 0;
 
    // Iterating through the array
    for (let 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
    let arr = [ 1, 2, 3, 4, 5 ];
    let N = arr.length;
 
    document.write(operation(arr, N));
 
</script>


Output

7

Time Complexity: O(N), where N is the length of the array.
Auxiliary Space: O(1) because constant space is being used for variables
 



Last Updated : 09 Feb, 2023
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