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 current element is even then, increment the value of L by 1.
- Otherwise, add the value L * (L + 1) / 2 to the variable count and update L to 0.
- 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 <bits/stdc++.h>using namespace std;// Function to count the number of// subarrays having even Bitwise ORint 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 Codeint main(){ int arr[] = { 1, 5, 4, 2, 6 }; int N = sizeof(arr) / sizeof(arr[0]); // Function Call cout << bitOr(arr, N); return 0;} |
Java
// Java program for the above approachimport java.util.*;class GFG{ // Function to count the number of// subarrays having even Bitwise ORstatic 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 Codepublic 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 ORdef 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 Codeif __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 approachusing 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
<script>// Javascript program for the above approach// Function to count the number of// subarrays having even Bitwise ORfunction bitOr(arr, N){ // Store number of subarrays // having even bitwise OR var count = 0; // Store the length of the current // subarray having even numbers var length = 0; var i; // Traverse the array for (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 += Math.floor((length) * (length + 1) / 2); } // Make length of subarray as 0 length = 0; } } // Add contribution of previous subarray count += Math.floor((length) * (length + 1) / 2); // Return total count of subarrays return count;}// Driver Code var arr = [1, 5, 4, 2, 6] var N = arr.length; // Function Call document.write(bitOr(arr, N));</script> |
Output:
6
Time Complexity: O(N)
Auxiliary Space: O(1)
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