Open In App

Sum of bitwise OR of all subarrays

Improve
Improve
Improve
Like Article
Like
Save Article
Save
Share
Report issue
Report

Given an array of positive integers, find the total sum after performing the bit wise OR operation on all the sub arrays of a given array.

Examples: 

Input : 1 2 3 4 5
Output : 71

Input : 6 5 4 3 2
Output : 84

First initialize the two variable sum=0, sum1=0, variable sum will store the total sum and, with sum1 we will perform bitwise OR operation for each jth element, and add sum1 with sum. 

  1. Traverse the from 0th position to n-1. 
  2. For each ith variable we will perform bit wise OR operation on all the sub arrays to find the total sum. 

Repeat step until the whole array is traverse. 

C++




// C++ program to find sum of
// bitwise ors of all subarrays.
#include <iostream>
using namespace std;
 
int totalSum(int a[], int n)
{
    int i, sum = 0, sum1 = 0, j;
 
    for (i = 0; i < n; i++)
    {
 
        sum1 = 0;
 
        // perform Bitwise OR operation
        // on all the subarray present
        // in array
        for (j = i; j < n; j++)
        {
 
            // OR operation
            sum1 = (sum1 | a[j]);
 
            // now add the sum after performing
            // the Bitwise OR operation
            sum = sum + sum1;
        }
    }
 
    return sum;
}
 
// Driver code
int main()
{
    int a[] = { 1, 2, 3, 4, 5 };
    int n = sizeof(a) / sizeof(a[0]);
    cout << totalSum(a, n) << endl;
    return 0;
}
 
// This code is contributed
// by Shivi_Aggarwal


C




// C program to find sum of bitwise ors
// of all subarrays.
#include <stdio.h>
 
int totalSum(int a[], int n)
{
    int i, sum = 0, sum1 = 0, j;
 
    for (i = 0; i < n; i++) {
 
        sum1 = 0;
 
        // perform Bitwise OR operation
        // on all the subarray present in array
        for (j = i; j < n; j++) {
 
            // OR operation
            sum1 = (sum1 | a[j]);
 
            // now add the sum after performing the
            // Bitwise OR operation
            sum = sum + sum1;
        }
    }
 
    return sum;
}
 
// Driver code
int main()
{
    int a[] = { 1, 2, 3, 4, 5 };
    int n = sizeof(a)/sizeof(a[0]);
    printf("%d ", totalSum(a, n));
    return 0;
}


Java




// Java program to find sum
// of bitwise ors of all subarrays.
import java.util.*;
import java.lang.*;
import java.io.*;
 
class GFG
{
static int totalSum(int a[], int n)
{
    int i, sum = 0, sum1 = 0, j;
 
    for (i = 0; i < n; i++)
    {
        sum1 = 0;
 
        // perform Bitwise OR operation
        // on all the subarray present
        // in array
        for (j = i; j < n; j++)
        {
 
            // OR operation
            sum1 = (sum1 | a[j]);
 
            // now add the sum after
            // performing the Bitwise
            // OR operation
            sum = sum + sum1;
        }
    }
 
    return sum;
}
 
// Driver code
public static void main(String args[])
{
    int a[] = { 1, 2, 3, 4, 5 };
    int n = a.length;
    System.out.println(totalSum(a,n));
}
}
 
// This code is contributed
// by Subhadeep


Python3




# Python3 program to find sum of
# bitwise ors of all subarrays.
def totalSum(a, n):
    sum = 0;
    for i in range(n):
        sum1 = 0;
         
        # perform Bitwise OR operation
        # on all the subarray present
        # in array
        for j in range(i, n):
             
            # OR operation
            sum1 = (sum1 | a[j]);
             
            # now add the sum after
            # performing the
            # Bitwise OR operation
            sum = sum + sum1;
    return sum;
 
# Driver code
a = [1, 2, 3, 4, 5];
n = len(a);
print(totalSum(a, n));
 
# This code is contributed by mits


C#




// C# program to find sum
// of bitwise ors of all
// subarrays.
using System;
 
class GFG
{
static int totalSum(int[] a, int n)
{
    int sum = 0;
    for(int i = 0; i < n; i++)
    {
        int sum1 = 0;
 
        // perform Bitwise OR operation
        // on all the subarray present
        // in array
        for (int j = i; j < n; j++)
        {
 
            // OR operation
            sum1 = (sum1 | a[j]);
 
            // now add the sum after
            // performing the Bitwise
            // OR operation
            sum = sum + sum1;
        }
    }
 
    return sum;
}
 
// Driver code
static void Main()
{
    int[] a = { 1, 2, 3, 4, 5 };
    int n = a.Length;
    Console.WriteLine(totalSum(a,n));
}
}
 
// This code is contributed
// by mits


PHP




<?php
// PHP program to find
// sum of bitwise ors
// of all subarrays.
function totalSum($a,$n)
{
$sum = 0;
for ($i = 0; $i < $n; $i++)
{
    $sum1 = 0;
 
    // perform Bitwise OR operation
    // on all the subarray present
    // in array
    for ($j = $i; $j < $n; $j++)
    {
 
        // OR operation
        $sum1 = ($sum1 | $a[$j]);
 
        // now add the sum after
        // performing the
        // Bitwise OR operation
        $sum = $sum + $sum1;
    }
}
return $sum;
}
 
// Driver code
$a = array(1, 2, 3, 4, 5);
$n = sizeof($a);
echo totalSum($a, $n);
 
// This code is contributed by mits
?>


Javascript




<script>
 
// Java program to find sum
// of bitwise ors of all subarrays.
function totalSum(a, n)
{
    let i, sum = 0, sum1 = 0, j;
 
    for (i = 0; i < n; i++)
    {
        sum1 = 0;
 
        // perform Bitwise OR operation
        // on all the subarray present
        // in array
        for (j = i; j < n; j++)
        {
 
            // OR operation
            sum1 = (sum1 | a[j]);
 
            // now add the sum after
            // performing the Bitwise
            // OR operation
            sum = sum + sum1;
        }
    }
    return sum;
}
 
// Driver code
    let a = [ 1, 2, 3, 4, 5 ];
    let n = a.length;
    document.write(totalSum(a,n));
 
// This code is contributed shivanisinghss2110
</script>


Output

71

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



Last Updated : 28 Feb, 2023
Like Article
Save Article
Previous
Next
Share your thoughts in the comments
Similar Reads