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.
- Traverse the from 0th position to n-1.
- 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 codeint 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 codeint 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 codepublic 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 codea = [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 codestatic 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)
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