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(N^{2})

**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); ` `} ` |

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## 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 ` |

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## 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 ` |

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## 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 ` |

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**Output:**

7

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

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