Sum of Bitwise OR of every array element paired with all other array elements
Given an array arr[] consisting of non-negative integers, the task for each array element arr[i] is to print the sum of Bitwise OR of all pairs (arr[i], arr[j]) ( 0 ≤ j ≤ N ).
Examples:
Input: arr[] = {1, 2, 3, 4}
Output: 12 14 16 22
Explanation:
For i = 0 the required sum will be (1 | 1) + (1 | 2) + (1 | 3) + (1 | 4) = 12
For i = 1 the required sum will be (2 | 1) + (2 | 2) + (2 | 3) + (2 | 4) = 14
For i = 2 the required sum will be (3 | 1) + (3 | 2) + (3 | 3) + (3 | 4) = 16
For i = 3 the required sum will be (4 | 1) + (4 | 2) + (4 | 3) + (4 | 4) = 22Input: arr[] = {3, 2, 5, 4, 8}
Output: 31 28 37 34 54
Naive Approach: The simplest approach for every array element arr[i] is to traverse the array and calculate sum of Bitwise OR of all possible (arr[i], arr[j]) and print the obtained sum.
Below is the implementation of the above approach:
C++
// C++ program for the above approach #include <bits/stdc++.h> using namespace std; // Function to print required sum for // every valid index i void printORSumforEachElement( int arr[], int N) { for ( int i = 0; i < N; i++) { // Store the required sum // for current array element int req_sum = 0; // Generate all possible pairs (arr[i], arr[j]) for ( int j = 0; j < N; j++) { // Update the value of req_sum req_sum += (arr[i] | arr[j]); } // Print the required sum cout << req_sum << " " ; } } // Driver Code int main() { // Given array int arr[] = { 1, 2, 3, 4 }; // Size of the array int N = sizeof (arr) / sizeof (arr[0]); // Function Call printORSumforEachElement(arr, N); return 0; } |
Java
// Java program for the above approach import java.util.*; class GFG { // Function to print required sum for // every valid index i static void printORSumforEachElement( int arr[], int N) { for ( int i = 0 ; i < N; i++) { // Store the required sum // for current array element int req_sum = 0 ; // Generate all possible pairs (arr[i], arr[j]) for ( int j = 0 ; j < N; j++) { // Update the value of req_sum req_sum += (arr[i] | arr[j]); } // Print the required sum System.out.print(req_sum+ " " ); } } // Driver Code public static void main(String[] args) { // Given array int arr[] = { 1 , 2 , 3 , 4 }; // Size of the array int N = arr.length; // Function Call printORSumforEachElement(arr, N); } } // This code is contributed by 29AjayKumar |
Python3
# Python3 program for the above approach # Function to print required sum for # every valid index i def printORSumforEachElement(arr, N): for i in range ( 0 , N): # Store the required sum # for current array element req_sum = 0 # Generate all possible pairs(arr[i],arr[j]) for j in range ( 0 , N): # Update the value of req_sum req_sum + = (arr[i] | arr[j]) # Print required sum print (req_sum, end = " " ) # Driver code # Given array arr = [ 1 , 2 , 3 , 4 ] # Size of array N = len (arr) # Function call printORSumforEachElement(arr, N) # This code is contributed by Virusbuddah |
C#
// C# program for the above approach using System; public class GFG { // Function to print required sum for // every valid index i static void printORSumforEachElement( int []arr, int N) { for ( int i = 0; i < N; i++) { // Store the required sum // for current array element int req_sum = 0; // Generate all possible pairs (arr[i], arr[j]) for ( int j = 0; j < N; j++) { // Update the value of req_sum req_sum += (arr[i] | arr[j]); } // Print the required sum Console.Write(req_sum+ " " ); } } // Driver Code public static void Main(String[] args) { // Given array int []arr = { 1, 2, 3, 4 }; // Size of the array int N = arr.Length; // Function Call printORSumforEachElement(arr, N); } } // This code is contributed by 29AjayKumar |
Javascript
<script> // Javascript program of the above approach // Function to print required sum for // every valid index i function prletORSumforEachElement(arr, N) { for (let i = 0; i < N; i++) { // Store the required sum // for current array element let req_sum = 0; // Generate all possible pairs // (arr[i], arr[j]) for (let j = 0; j < N; j++) { // Update the value of req_sum req_sum += (arr[i] | arr[j]); } // Print the required sum document.write(req_sum + " " ); } } // Driver Code // Given array let arr = [ 1, 2, 3, 4 ]; // Size of the array let N = arr.length; // Function Call prletORSumforEachElement(arr, N); // This code is contributed by avijitmondal1998 </script> |
12 14 16 22
Time Complexity: O(N2)
Auxiliary Space: O(1)
Efficient Approach: The optimal idea is to use Bit Manipulation by using the assumption that integers are represented using 32 bits. Follow the steps below to solve the problem:
- Consider every k-th bit and use a frequency array freq[] to store the count of elements for which k-th bit is set.
- For every array element check whether k-th bit of that element is set or not. If k-th bit is set then simply increase the frequency of k-th bit.
- Traverse the array and for every element arr[i] check whether k-th bit of arr[i] is set or not.
- Initialize required sum to 0 for every index i.
- If k-th bit of arr[i] is set then it means, k-th bit of every possible (arr[i] | arr[j]) will also be set. So in this case add (1 << k) * N to the required sum.
- Otherwise, if k-th bit of arr[i] is not set then it means that k-th bit of (arr[i] | arr[j]) will be set if and only if k-th bit of arr[j] is set. So in this case add (1 << k) * freq[k] to the required sum which is previously calculated that k-th bit is set for freq[k] number of elements.
- Finally, print the value of required sum for index i.
Below is the implementation of the above approach:
C++
// C++ program for the above approach #include <bits/stdc++.h> using namespace std; // Function to print required sum for // every valid index i void printORSumforEachElement( int arr[], int N) { // Frequency array to store frequency // of every k-th bit int freq[32]; // Initialize frequency array memset (freq, 0, sizeof freq); for ( int i = 0; i < N; i++) { for ( int k = 0; k < 32; k++) { // If k-th bit is set, then update // the frequency of k-th bit if ((arr[i] & (1 << k)) != 0) freq[k]++; } } for ( int i = 0; i < N; i++) { // Stores the required sum // for the current array element int req_sum = 0; for ( int k = 0; k < 32; k++) { // If k-th bit is set if ((arr[i] & (1 << k)) != 0) req_sum += (1 << k) * N; // If k-th bit is not set else req_sum += (1 << k) * freq[k]; } // Print the required sum cout << req_sum << " " ; } } // Driver Code int main() { // Given array int arr[] = { 1, 2, 3, 4 }; // Size of the array int N = sizeof (arr) / sizeof (arr[0]); // Function Call printORSumforEachElement(arr, N); return 0; } |
Java
// Java program for the above approach import java.util.*; class GFG { // Function to print required sum for // every valid index i static void printORSumforEachElement( int arr[], int N) { // Frequency array to store frequency // of every k-th bit int []freq = new int [ 32 ]; for ( int i = 0 ; i < N; i++) { for ( int k = 0 ; k < 32 ; k++) { // If k-th bit is set, then update // the frequency of k-th bit if ((arr[i] & ( 1 << k)) != 0 ) freq[k]++; } } for ( int i = 0 ; i < N; i++) { // Stores the required sum // for the current array element int req_sum = 0 ; for ( int k = 0 ; k < 32 ; k++) { // If k-th bit is set if ((arr[i] & ( 1 << k)) != 0 ) req_sum += ( 1 << k) * N; // If k-th bit is not set else req_sum += ( 1 << k) * freq[k]; } // Print the required sum System.out.print(req_sum+ " " ); } } // Driver Code public static void main(String[] args) { // Given array int arr[] = { 1 , 2 , 3 , 4 }; // Size of the array int N = arr.length; // Function Call printORSumforEachElement(arr, N); } } // This code is contributed by shikhasingrajput |
Python3
# Python3 program for the above approach # Function to print required sum for # every valid index i def printORSumforEachElement(arr, N): # Frequency array to store frequency # of every k-th bit freq = [ 0 for i in range ( 32 )] for i in range (N): for k in range ( 32 ): # If k-th bit is set, then update # the frequency of k-th bit if ((arr[i] & ( 1 << k)) ! = 0 ): freq[k] + = 1 for i in range (N): # Stores the required sum # for the current array element req_sum = 0 for k in range ( 32 ): # If k-th bit is set if ((arr[i] & ( 1 << k)) ! = 0 ): req_sum + = ( 1 << k) * N # If k-th bit is not set else : req_sum + = ( 1 << k) * freq[k] # Print the required sum print (req_sum, end = " " ) # Driver Code if __name__ = = '__main__' : # Given array arr = [ 1 , 2 , 3 , 4 ] # Size of the array N = len (arr) # Function Call printORSumforEachElement(arr, N) # This code is contributed by SURENDRA_GANGWAR |
C#
// C# program for the above approach using System; class GFG { // Function to print required sum for // every valid index i static void printORSumforEachElement( int [] arr, int N) { // Frequency array to store frequency // of every k-th bit int [] freq = new int [32]; for ( int i = 0; i < N; i++) { for ( int k = 0; k < 32; k++) { // If k-th bit is set, then update // the frequency of k-th bit if ((arr[i] & (1 << k)) != 0) freq[k]++; } } for ( int i = 0; i < N; i++) { // Stores the required sum // for the current array element int req_sum = 0; for ( int k = 0; k < 32; k++) { // If k-th bit is set if ((arr[i] & (1 << k)) != 0) req_sum += (1 << k) * N; // If k-th bit is not set else req_sum += (1 << k) * freq[k]; } // Print the required sum Console.Write(req_sum + " " ); } } // Driver Code public static void Main() { // Given array int [] arr = { 1, 2, 3, 4 }; // Size of the array int N = arr.Length; // Function Call printORSumforEachElement(arr, N); } } // This code is contributed by chitranayal. |
Javascript
<script> // Javascript program for the above approach // Function to print required sum for // every valid index i function printORSumforEachElement(arr, N) { // Frequency array to store frequency // of every k-th bit var freq = Array(32).fill(0); for ( var i = 0; i < N; i++) { for ( var k = 0; k < 32; k++) { // If k-th bit is set, then update // the frequency of k-th bit if ((arr[i] & (1 << k)) != 0) freq[k]++; } } for ( var i = 0; i < N; i++) { // Stores the required sum // for the current array element var req_sum = 0; for ( var k = 0; k < 32; k++) { // If k-th bit is set if ((arr[i] & (1 << k)) != 0) req_sum += (1 << k) * N; // If k-th bit is not set else req_sum += (1 << k) * freq[k]; } // Print the required sum document.write(req_sum + " " ); } } // Driver Code // Given array var arr = [ 1, 2, 3, 4 ]; // Size of the array var N = arr.length; // Function Call printORSumforEachElement(arr, N); // This code is contributed by rutvik_56 </script> |
12 14 16 22
Time Complexity: O(32 * N)
Auxiliary Space: O(m), where m = 32
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