Sum of Bitwise OR of every array element paired with all other array elements
Last Updated :
15 Nov, 2021
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++
// 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 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 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# 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
<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>
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 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 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# 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.
<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>
Time Complexity: O(32 * N)
Auxiliary Space: O(m), where m = 32
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