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++
// 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> |
7
Time Complexity: O(N), where N is the length of the array.