# Count subarrays having even Bitwise OR

• Difficulty Level : Medium
• Last Updated : 22 Apr, 2021

Given an array arr[] consisting of N positive integers, the task is to count the number of subarrays whose Bitwise OR of its elements even.

Examples:

Input: arr[] = {1, 5, 4, 2, 6 }
Output: 6
Explanation:
The subarrays with even Bitwise OR are {4}, {2}, {6}, {2, 6}, {4, 2}, {4, 2, 6}.
Therefore, the number of subarrays having even Bitwise OR are 6.

Input: arr[] ={2, 5, 6, 8}
Output: 4

Naive Approach: The simplest approach to solve the problem is to generate all subarrays and if the Bitwise OR of any subarray is even, then increase the count of such subarrays. After checking for all the subarrays, print the count obtained as the result.

Time Complexity: O(N3)
Auxiliary Space: O(1)

Efficient Approach: The above approach can be optimized by observing the fact that if any of the elements in the subarray is odd, then it will surely make the Bitwise OR of the subarray odd. Therefore, the idea is to find the length of the continuous segment in the array which is even, and add its contribution to the total count.
Follow the steps below to solve the problem:

• Initialize a variable, say, count, to store the total number of subarrays having Bitwise OR as even.
• Initialize a variable, say L, to store the count of adjacent elements that are even.
• Traverse the given array arr[] and perform the following steps:
• If the value of L is non-zero, then add L*(L + 1)/2 to the variable count.
• After completing the above steps, print the value of 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 OR``int` `bitOr(``int` `arr[], ``int` `N)``{``    ``// Store number of subarrays``    ``// having even bitwise OR``    ``int` `count = 0;` `    ``// Store the length of the current``    ``// subarray having even numbers``    ``int` `length = 0;` `    ``// Traverse the array``    ``for` `(``int` `i = 0; i < N; i++) {` `        ``// If the element is even``        ``if` `(arr[i] % 2 == 0) {` `            ``// Increment size of the``            ``// current continuous sequence``// of even array elements``            ``length++;``        ``}` `        ``// If arr[i] is odd``        ``else` `{` `            ``// If length is non zero``            ``if` `(length != 0) {` `                ``// Adding contribution of``                ``// subarrays consisting``                ``// only of even numbers``                ``count += ((length)``                          ``* (length + 1))``                         ``/ 2;``            ``}` `            ``// Make length of subarray as 0``            ``length = 0;``        ``}``    ``}` `    ``// Add contribution of previous subarray``    ``count += ((length) * (length + 1)) / 2;` `    ``// Return total count of subarrays``    ``return` `count;``}` `// Driver Code``int` `main()``{``    ``int` `arr[] = { 1, 5, 4, 2, 6 };``    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr);` `    ``// Function Call``    ``cout << bitOr(arr, N);` `    ``return` `0;``}`

## Java

 `// Java program for the above approach``import` `java.util.*;` `class` `GFG{``    ` `// Function to count the number of``// subarrays having even Bitwise OR``static` `int` `bitOr(``int` `arr[], ``int` `N)``{``    ` `    ``// Store number of subarrays``    ``// having even bitwise OR``    ``int` `count = ``0``;` `    ``// Store the length of the current``    ``// subarray having even numbers``    ``int` `length = ``0``;` `    ``// Traverse the array``    ``for``(``int` `i = ``0``; i < N; i++)``    ``{``        ` `        ``// If the element is even``        ``if` `(arr[i] % ``2` `== ``0``)``        ``{``            ` `            ``// Increment size of the``            ``// current continuous sequence``            ``// of even array elements``            ``length++;``        ``}` `        ``// If arr[i] is odd``        ``else``        ``{``            ` `            ``// If length is non zero``            ``if` `(length != ``0``)``            ``{``                ` `                ``// Adding contribution of``                ``// subarrays consisting``                ``// only of even numbers``                ``count += ((length) * (length + ``1``)) / ``2``;``            ``}` `            ``// Make length of subarray as 0``            ``length = ``0``;``        ``}``    ``}` `    ``// Add contribution of previous subarray``    ``count += ((length) * (length + ``1``)) / ``2``;` `    ``// Return total count of subarrays``    ``return` `count;``}` `// Driver Code``public` `static` `void` `main(String[] args)``{``    ``int` `arr[] = { ``1``, ``5``, ``4``, ``2``, ``6` `};``    ``int` `N = arr.length;` `    ``// Function Call``    ``System.out.print(bitOr(arr, N));``}``}` `// This code is contributed by splevel62`

## Python3

 `# Python3 program for the above approach` `# Function to count the number of``# subarrays having even Bitwise OR``def` `bitOr(arr, N):``    ` `    ``# Store number of subarrays``    ``# having even bitwise OR``    ``count ``=` `0` `    ``# Store the length of the current``    ``# subarray having even numbers``    ``length ``=` `0` `    ``# Traverse the array``    ``for` `i ``in` `range``(N):` `        ``# If the element is even``        ``if` `(arr[i] ``%` `2` `=``=` `0``):` `            ``# Increment size of the``            ``# current continuous sequence``            ``# of even array elements``            ``length ``+``=` `1``            ` `        ``# If arr[i] is odd``        ``else``:``            ` `            ``# If length is non zero``            ``if` `(length !``=` `0``):` `                ``# Adding contribution of``                ``# subarrays consisting``                ``# only of even numbers``                ``count ``+``=` `((length) ``*` `(length ``+` `1``)) ``/``/` `2` `            ``# Make length of subarray as 0``            ``length ``=` `0` `    ``# Add contribution of previous subarray``    ``count ``+``=` `((length) ``*` `(length ``+` `1``)) ``/``/` `2` `    ``# Return total count of subarrays``    ``return` `count` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``    ` `    ``arr ``=` `[ ``1``, ``5``, ``4``, ``2``, ``6` `]``    ``N ``=` `len``(arr)` `    ``# Function Call``    ``print` `(bitOr(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 OR``  ``static` `int` `bitOr(``int``[] arr, ``int` `N)``  ``{` `    ``// Store number of subarrays``    ``// having even bitwise OR``    ``int` `count = 0;` `    ``// Store the length of the current``    ``// subarray having even numbers``    ``int` `length = 0;` `    ``// Traverse the array``    ``for``(``int` `i = 0; i < N; i++)``    ``{` `      ``// If the element is even``      ``if` `(arr[i] % 2 == 0)``      ``{` `        ``// Increment size of the``        ``// current continuous sequence``        ``// of even array elements``        ``length++;``      ``}` `      ``// If arr[i] is odd``      ``else``      ``{` `        ``// If length is non zero``        ``if` `(length != 0)``        ``{` `          ``// Adding contribution of``          ``// subarrays consisting``          ``// only of even numbers``          ``count += ((length) * (length + 1)) / 2;``        ``}` `        ``// Make length of subarray as 0``        ``length = 0;``      ``}``    ``}` `    ``// Add contribution of previous subarray``    ``count += ((length) * (length + 1)) / 2;` `    ``// Return total count of subarrays``    ``return` `count;``  ``}` `  ``// Driver code``  ``static` `void` `Main()``  ``{``    ``int``[] arr = { 1, 5, 4, 2, 6 };``    ``int` `N = arr.Length;` `    ``// Function Call``    ``Console.Write(bitOr(arr, N));``  ``}``}` `// This code is contributed by sanjoy_62.`

## Javascript

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Output:

`6`

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

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