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) = 22
Input: 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++
#include <bits/stdc++.h>
using namespace std;
void printORSumforEachElement( int arr[], int N)
{
for ( int i = 0; i < N; i++) {
int req_sum = 0;
for ( int j = 0; j < N; j++) {
req_sum += (arr[i] | arr[j]);
}
cout << req_sum << " " ;
}
}
int main()
{
int arr[] = { 1, 2, 3, 4 };
int N = sizeof (arr) / sizeof (arr[0]);
printORSumforEachElement(arr, N);
return 0;
}
|
Java
import java.util.*;
class GFG
{
static void printORSumforEachElement( int arr[], int N)
{
for ( int i = 0 ; i < N; i++)
{
int req_sum = 0 ;
for ( int j = 0 ; j < N; j++)
{
req_sum += (arr[i] | arr[j]);
}
System.out.print(req_sum+ " " );
}
}
public static void main(String[] args)
{
int arr[] = { 1 , 2 , 3 , 4 };
int N = arr.length;
printORSumforEachElement(arr, N);
}
}
|
Python3
def printORSumforEachElement(arr, N):
for i in range ( 0 , N):
req_sum = 0
for j in range ( 0 , N):
req_sum + = (arr[i] | arr[j])
print (req_sum, end = " " )
arr = [ 1 , 2 , 3 , 4 ]
N = len (arr)
printORSumforEachElement(arr, N)
|
C#
using System;
public class GFG
{
static void printORSumforEachElement( int []arr, int N)
{
for ( int i = 0; i < N; i++)
{
int req_sum = 0;
for ( int j = 0; j < N; j++)
{
req_sum += (arr[i] | arr[j]);
}
Console.Write(req_sum+ " " );
}
}
public static void Main(String[] args)
{
int []arr = { 1, 2, 3, 4 };
int N = arr.Length;
printORSumforEachElement(arr, N);
}
}
|
Javascript
<script>
function prletORSumforEachElement(arr, N)
{
for (let i = 0; i < N; i++)
{
let req_sum = 0;
for (let j = 0; j < N; j++)
{
req_sum += (arr[i] | arr[j]);
}
document.write(req_sum + " " );
}
}
let arr = [ 1, 2, 3, 4 ];
let N = arr.length;
prletORSumforEachElement(arr, N);
</script>
|
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++
#include <bits/stdc++.h>
using namespace std;
void printORSumforEachElement( int arr[], int N)
{
int freq[32];
memset (freq, 0, sizeof freq);
for ( int i = 0; i < N; i++) {
for ( int k = 0; k < 32; k++) {
if ((arr[i] & (1 << k)) != 0)
freq[k]++;
}
}
for ( int i = 0; i < N; i++) {
int req_sum = 0;
for ( int k = 0; k < 32; k++) {
if ((arr[i] & (1 << k)) != 0)
req_sum += (1 << k) * N;
else
req_sum += (1 << k) * freq[k];
}
cout << req_sum << " " ;
}
}
int main()
{
int arr[] = { 1, 2, 3, 4 };
int N = sizeof (arr) / sizeof (arr[0]);
printORSumforEachElement(arr, N);
return 0;
}
|
Java
import java.util.*;
class GFG
{
static void printORSumforEachElement( int arr[], int N)
{
int []freq = new int [ 32 ];
for ( int i = 0 ; i < N; i++)
{
for ( int k = 0 ; k < 32 ; k++)
{
if ((arr[i] & ( 1 << k)) != 0 )
freq[k]++;
}
}
for ( int i = 0 ; i < N; i++)
{
int req_sum = 0 ;
for ( int k = 0 ; k < 32 ; k++)
{
if ((arr[i] & ( 1 << k)) != 0 )
req_sum += ( 1 << k) * N;
else
req_sum += ( 1 << k) * freq[k];
}
System.out.print(req_sum+ " " );
}
}
public static void main(String[] args)
{
int arr[] = { 1 , 2 , 3 , 4 };
int N = arr.length;
printORSumforEachElement(arr, N);
}
}
|
Python3
def printORSumforEachElement(arr, N):
freq = [ 0 for i in range ( 32 )]
for i in range (N):
for k in range ( 32 ):
if ((arr[i] & ( 1 << k)) ! = 0 ):
freq[k] + = 1
for i in range (N):
req_sum = 0
for k in range ( 32 ):
if ((arr[i] & ( 1 << k)) ! = 0 ):
req_sum + = ( 1 << k) * N
else :
req_sum + = ( 1 << k) * freq[k]
print (req_sum, end = " " )
if __name__ = = '__main__' :
arr = [ 1 , 2 , 3 , 4 ]
N = len (arr)
printORSumforEachElement(arr, N)
|
C#
using System;
class GFG
{
static void printORSumforEachElement( int [] arr, int N)
{
int [] freq = new int [32];
for ( int i = 0; i < N; i++)
{
for ( int k = 0; k < 32; k++)
{
if ((arr[i] & (1 << k)) != 0)
freq[k]++;
}
}
for ( int i = 0; i < N; i++)
{
int req_sum = 0;
for ( int k = 0; k < 32; k++)
{
if ((arr[i] & (1 << k)) != 0)
req_sum += (1 << k) * N;
else
req_sum += (1 << k) * freq[k];
}
Console.Write(req_sum + " " );
}
}
public static void Main()
{
int [] arr = { 1, 2, 3, 4 };
int N = arr.Length;
printORSumforEachElement(arr, N);
}
}
|
Javascript
<script>
function printORSumforEachElement(arr, N)
{
var freq = Array(32).fill(0);
for ( var i = 0; i < N; i++)
{
for ( var k = 0; k < 32; k++)
{
if ((arr[i] & (1 << k)) != 0)
freq[k]++;
}
}
for ( var i = 0; i < N; i++)
{
var req_sum = 0;
for ( var k = 0; k < 32; k++)
{
if ((arr[i] & (1 << k)) != 0)
req_sum += (1 << k) * N;
else
req_sum += (1 << k) * freq[k];
}
document.write(req_sum + " " );
}
}
var arr = [ 1, 2, 3, 4 ];
var N = arr.length;
printORSumforEachElement(arr, N);
</script>
|
Time Complexity: O(32 * N)
Auxiliary Space: O(m), where m = 32
Last Updated :
15 Nov, 2021
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