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Sum of bitwise OR of all subarrays

  • Difficulty Level : Easy
  • Last Updated : 22 Jun, 2021

Give 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

 


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