# Count subarrays having even Bitwise XOR

• Difficulty Level : Hard
• Last Updated : 05 Jul, 2022

Given an array arr[] of size N, the task is to count the number of subarrays from the given array whose Bitwise XOR  is even.

Examples:

Input: arr[] = {1, 2, 3, 4}
Output: 4
Explanation: The subarrays having even Bitwise XOR are {{2}, {4}, {1, 2, 3}, {1, 2, 3, 4}}.

Input: arr[] = {2, 4, 6}
Output: 6
Explanation: The subarrays having even Bitwise XOR are {{2}, {4}, {6}, {2, 4}, {4, 6}, {2, 4, 6}}.

Naive Approach: The simplest approach to solve this problem is to generate all possible subarrays and check for every subarray, whether Bitwise XOR of all its elements are even or not. If found to be even, then increase the count. Finally, print the count as the result.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach``#include ``using` `namespace` `std;` `// Function to count the number of``// subarrays having even Bitwise XOR``void` `evenXorSubarray(``int` `arr[], ``int` `n)``{``    ``// Store the required result``    ``int` `ans = 0;` `    ``// Generate subarrays with``    ``// arr[i] as the first element``    ``for` `(``int` `i = 0; i < n; i++) {` `        ``// Store XOR of current subarray``        ``int` `XOR = 0;` `        ``// Generate subarrays with``        ``// arr[j] as the last element``        ``for` `(``int` `j = i; j < n; j++) {` `            ``// Calculate Bitwise XOR``            ``// of the current subarray``            ``XOR = XOR ^ arr[j];` `            ``// If XOR is even,``            ``// increase ans by 1``            ``if` `((XOR & 1) == 0)``                ``ans++;``        ``}``    ``}` `    ``// Print the result``    ``cout << ans;``}` `// Driver Code``int` `main()``{``    ``// Given array``    ``int` `arr[] = { 1, 2, 3, 4 };` `    ``// Stores the size of the array``    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr[0]);` `    ``// Function Call``    ``evenXorSubarray(arr, N);` `    ``return` `0;``}`

## Java

 `// Java program for the above approach``import` `java.io.*;``import` `java.lang.*;``import` `java.util.*;` `class` `GFG {` `  ``// Function to count the number of``  ``// subarrays having even Bitwise XOR``  ``static` `void` `evenXorSubarray(``int` `arr[], ``int` `n)``  ``{``    ` `    ``// Store the required result``    ``int` `ans = ``0``;` `    ``// Generate subarrays with``    ``// arr[i] as the first element``    ``for` `(``int` `i = ``0``; i < n; i++) {` `      ``// Store XOR of current subarray``      ``int` `XOR = ``0``;` `      ``// Generate subarrays with``      ``// arr[j] as the last element``      ``for` `(``int` `j = i; j < n; j++) {` `        ``// Calculate Bitwise XOR``        ``// of the current subarray``        ``XOR = XOR ^ arr[j];` `        ``// If XOR is even,``        ``// increase ans by 1``        ``if` `((XOR & ``1``) == ``0``)``          ``ans++;``      ``}``    ``}` `    ``// Print the result``    ``System.out.println(ans);``  ``}` `  ``// Driver Code``  ``public` `static` `void` `main(String[] args)``  ``{` `    ``// Given array``    ``int` `arr[] = { ``1``, ``2``, ``3``, ``4` `};` `    ``// Stores the size of the array``    ``int` `N = arr.length;``    ``evenXorSubarray(arr, N);``  ``}``}` `// This code is contributed by Kingash.`

## Python3

 `# Python3 program for the above approach` `# Function to count the number of``# subarrays having even Bitwise XOR``def` `evenXorSubarray(arr, n):``  ` `    ``# Store the required result``    ``ans ``=` `0` `    ``# Generate subarrays with``    ``# arr[i] as the first element``    ``for` `i ``in` `range``(n):` `        ``# Store XOR of current subarray``        ``XOR ``=` `0` `        ``# Generate subarrays with``        ``# arr[j] as the last element``        ``for` `j ``in` `range``(i, n):` `            ``# Calculate Bitwise XOR``            ``# of the current subarray``            ``XOR ``=` `XOR ^ arr[j]` `            ``# If XOR is even,``            ``# increase ans by 1``            ``if` `((XOR & ``1``) ``=``=` `0``):``                ``ans ``+``=` `1` `    ``# Print the result``    ``print` `(ans)` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``  ` `    ``# Given array``    ``arr ``=` `[``1``, ``2``, ``3``, ``4``]` `    ``# Stores the size of the array``    ``N ``=` `len``(arr)` `    ``# Function Call``    ``evenXorSubarray(arr, N)` `# This code is contributed by mohit kumar 29.`

## C#

 `// C# program for the above approach``using` `System;``class` `GFG``{``  ``// Function to count the number of``  ``// subarrays having even Bitwise XOR``  ``static` `void` `evenXorSubarray(``int``[] arr, ``int` `n)``  ``{` `    ``// Store the required result``    ``int` `ans = 0;` `    ``// Generate subarrays with``    ``// arr[i] as the first element``    ``for` `(``int` `i = 0; i < n; i++) {` `      ``// Store XOR of current subarray``      ``int` `XOR = 0;` `      ``// Generate subarrays with``      ``// arr[j] as the last element``      ``for` `(``int` `j = i; j < n; j++) {` `        ``// Calculate Bitwise XOR``        ``// of the current subarray``        ``XOR = XOR ^ arr[j];` `        ``// If XOR is even,``        ``// increase ans by 1``        ``if` `((XOR & 1) == 0)``          ``ans++;``      ``}``    ``}` `    ``// Print the result``    ``Console.WriteLine(ans);``  ``}` `  ``// Driver Code``  ``public` `static` `void` `Main()``  ``{``    ``// Given array``    ``int``[] arr = { 1, 2, 3, 4 };` `    ``// Stores the size of the array``    ``int` `N = arr.Length;``    ``evenXorSubarray(arr, N);``  ``}``}` `// This code is contributed by souravghosh0416.`

## Javascript

 ``

Output:

`4`

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

Efficient Approach: The above approach can be optimized based on following observations:

Bitwise XOR of all elements in the range of indices [i + 1, j] be A.
Bitwise XOR of all elements in the range of indices [0, i] be B.
Bitwise XOR of all elements in the range of indices [0, j] be C
By performing B ^ C,  the common elements from the two ranges cancel out, resulting in XOR of all elements from the range [i + 1, j]

Therefore, the idea is to update the number of subarrays starting from index 0, having even and odd XOR values while traversing the array and updating the count accordingly. Follow the steps below to solve the problem:

• Initialize two variables, say ans and XOR, to store the required count of subarrays and store XOR values of subarrays respectively.
• Initialize an array, say freq[] of size 2, where freq[0] denotes the number of subarrays having even XOR value, and freq[1] denotes the number of subarrays having odd XOR values, starting from the index 0.
• Traverse the array, arr[] using a variable, say i, and for each array element:
• Update the value of XOR to XOR ^ arr[i].
• If XOR is even, then increase the value of ans by 1 + freq[0].
• Increment freq[0] by 1, because when subarray {arr[0], i] is XORed with other subarrays having even xor value then it would result in an even XOR value. Also, 1 is added to consider the entire subarray[0, i].
• Similarly, if XOR is odd, increment ans by freq[1] and increment freq[1] by 1.
• After completing the above steps, print the value of ans as the result.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach``#include ``using` `namespace` `std;` `// Function to count subarrays``// having even Bitwise XOR``void` `evenXorSubarray(``int` `arr[], ``int` `n)``{``    ``// Store the required result``    ``int` `ans = 0;` `    ``// Stores count of subarrays``    ``// with even and odd XOR values``    ``int` `freq[] = { 0, 0 };` `    ``// Stores Bitwise XOR of``    ``// current subarray``    ``int` `XOR = 0;` `    ``// Traverse the array``    ``for` `(``int` `i = 0; i < n; i++) {` `        ``// Update current Xor``        ``XOR = XOR ^ arr[i];` `        ``// If XOR is even``        ``if` `(XOR % 2 == 0) {` `            ``// Update ans``            ``ans += freq[0] + 1;` `            ``// Increment count of``            ``// subarrays with even XOR``            ``freq[0]++;``        ``}``        ``else` `{` `            ``// Otherwise, increment count``            ``// of subarrays with odd XOR``            ``ans += freq[1];``            ``freq[1]++;``        ``}``    ``}` `    ``// Print the result``    ``cout << ans;``}` `// Driver Code``int` `main()``{``    ``// Given array``    ``int` `arr[] = { 1, 2, 3, 4 };` `    ``// Stores the size of the array``    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr[0]);` `    ``evenXorSubarray(arr, N);` `    ``return` `0;``}`

## Java

 `// Java program for the above approach``import` `java.io.*;``import` `java.lang.*;``import` `java.util.*;` `class` `GFG {` `    ``// Function to count subarrays``    ``// having even Bitwise XOR``    ``static` `void` `evenXorSubarray(``int` `arr[], ``int` `n)``    ``{``        ``// Store the required result``        ``int` `ans = ``0``;` `        ``// Stores count of subarrays``        ``// with even and odd XOR values``        ``int` `freq[] = { ``0``, ``0` `};` `        ``// Stores Bitwise XOR of``        ``// current subarray``        ``int` `XOR = ``0``;` `        ``// Traverse the array``        ``for` `(``int` `i = ``0``; i < n; i++) {` `            ``// Update current Xor``            ``XOR = XOR ^ arr[i];` `            ``// If XOR is even``            ``if` `(XOR % ``2` `== ``0``) {` `                ``// Update ans``                ``ans += freq[``0``] + ``1``;` `                ``// Increment count of``                ``// subarrays with even XOR``                ``freq[``0``]++;``            ``}``            ``else` `{` `                ``// Otherwise, increment count``                ``// of subarrays with odd XOR``                ``ans += freq[``1``];``                ``freq[``1``]++;``            ``}``        ``}` `        ``// Print the result``        ``System.out.println(ans);``    ``}` `    ``// Driver Code``    ``public` `static` `void` `main(String[] args)``    ``{` `        ``// Given array``        ``int` `arr[] = { ``1``, ``2``, ``3``, ``4` `};` `        ``// Stores the size of the array``        ``int` `N = arr.length;` `        ``evenXorSubarray(arr, N);``    ``}``}`

## Python3

 `# Python3 program for the above approach` `# Function to count subarrays``# having even Bitwise XOR``def` `evenXorSubarray(arr, n):``    ` `    ``# Store the required result``    ``ans ``=` `0` `    ``# Stores count of subarrays``    ``# with even and odd XOR values``    ``freq ``=` `[``0``] ``*` `n` `    ``# Stores Bitwise XOR of``    ``# current subarray``    ``XOR ``=` `0` `    ``# Traverse the array``    ``for` `i ``in` `range``(n):` `        ``# Update current Xor``        ``XOR ``=` `XOR ^ arr[i]` `        ``# If XOR is even``        ``if` `(XOR ``%` `2` `=``=` `0``):` `            ``# Update ans``            ``ans ``+``=` `freq[``0``] ``+` `1` `            ``# Increment count of``            ``# subarrays with even XOR``            ``freq[``0``] ``+``=` `1``        ` `        ``else``:` `            ``# Otherwise, increment count``            ``# of subarrays with odd XOR``            ``ans ``+``=` `freq[``1``]``            ``freq[``1``] ``+``=` `1``        ` `    ``# Print the result``    ``print``(ans)` `# Driver Code` `# Given array``arr ``=` `[ ``1``, ``2``, ``3``, ``4` `]` `# Stores the size of the array``N ``=` `len``(arr)` `evenXorSubarray(arr, N)` `# This code is contributed by sanjoy_62`

## C#

 `// C# program for the above approach``using` `System;``class` `GFG``{``  ``// Function to count subarrays``  ``// having even Bitwise XOR``  ``static` `void` `evenXorSubarray(``int``[] arr, ``int` `n)``  ``{``    ` `    ``// Store the required result``    ``int` `ans = 0;` `    ``// Stores count of subarrays``    ``// with even and odd XOR values``    ``int``[] freq = { 0, 0 };` `    ``// Stores Bitwise XOR of``    ``// current subarray``    ``int` `XOR = 0;` `    ``// Traverse the array``    ``for` `(``int` `i = 0; i < n; i++) {` `      ``// Update current Xor``      ``XOR = XOR ^ arr[i];` `      ``// If XOR is even``      ``if` `(XOR % 2 == 0) {` `        ``// Update ans``        ``ans += freq[0] + 1;` `        ``// Increment count of``        ``// subarrays with even XOR``        ``freq[0]++;``      ``}``      ``else` `{` `        ``// Otherwise, increment count``        ``// of subarrays with odd XOR``        ``ans += freq[1];``        ``freq[1]++;``      ``}``    ``}` `    ``// Print the result``    ``Console.WriteLine(ans);``  ``}`  `  ``// Driver Code``  ``public` `static` `void` `Main(String[] args)``  ``{``    ``// Given array``    ``int``[] arr = { 1, 2, 3, 4 };` `    ``// Stores the size of the array``    ``int` `N = arr.Length;` `    ``evenXorSubarray(arr, N);``  ``}``}` `// This code is contributed by susmitakundugoaldanga.`

## Javascript

 ``

Output:

`4`

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

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