Find bitwise OR of all possible sub-arrays

Difficulty Level : Easy
Last Updated : 11 May, 2021

Given an array A of size N where, $1\leq N \leq 10^{5}$ . The task is to find the OR of all possible sub-arrays of A and then the OR of all these results.
Examples:

Input : 1 4 6
Output : 7
All possible subarrays are 
{1}, {1, 4}, {4, 6} and {1, 4, 6}
ORs of these subarrays are 1, 5, 6
and 7. OR of these ORs is 7.

Input : 10 100 1000
Output : 1006

Approach: In SET 1 we have seen how to find bitwise AND of all possible subarrays. A similar approach is also applicable here.
The Naive solution is to find the OR of all the sub-arrays and then output the OR of their results. This will lead to O(N²) solution.
Efficient Solution: Using the property that $X\|X\|...\|X=X$ i:e it doesn’t matter how many times an element comes, it’s ORing will be counted as one only. Thus our problem boils down to finding the OR of all the elements of the array.
Below is the implementation of above approach.

C++

// C++ program to find OR of all the sub-arrays
#include <bits/stdc++.h>
using namespace std;
 
// function to return OR of sub-arrays
int OR(int a[], int n)
{
    int ans = a[0];
    for (int i = 1; i < n; ++i)
         ans |= a[i];   
 
    return ans;
}
 
// Driver program
int main()
{
    int a[] = { 1, 4, 6 };
    int n = sizeof(a) / sizeof(a[0]);
 
    // print OR of all subarrays
    cout << OR(a, n);
 
    return 0;
}

Java

// Java program to find OR of
// all the sub-arrays
class GFG
{
     
// function to return OR
// of sub-arrays
static int OR(int a[],int n)
{
    int ans = a[0];
    int i;
    for(i = 1; i < n; i++)
    {
        ans |= a[i];
    }
    return ans;
}
 
// Driver Code
public static void main(String args[])
{
    int a[] = { 1, 4, 6};
     
    int n = a.length;
     
    // print OR of all subarrays
    System.out.println(OR(a, n));
}
}
 
// This code is contributed
// by ANKITRAI1

Python3

# Python3 program to find OR of all the sub-arrays
 
# function to return OR of sub-arrays
def OR(a, n):
     
    ans = a[0]
    for i in range(1,n):
        ans |= a[i]
     
    return ans
     
# Driver Code
if __name__=='__main__':
    a = [1, 4, 6]
    n = len(a)
 
# print OR of all subarrays
    print(OR(a, n))
 
# This code is contributed
# by Shashank_Sharma

C#

// C# program to find OR of
// all the sub-arrays
using System;
 
class GFG
{
     
// function to return OR
// of sub-arrays
static int OR(int[] a, int n)
{
    int ans = a[0];
    int i;
    for(i = 1; i < n; i++)
    {
        ans |= a[i];
    }
    return ans;
}
 
// Driver Code
public static void Main()
{
    int[] a = { 1, 4, 6};
     
    int n = a.Length;
     
    // print OR of all subarrays
    Console.Write(OR(a, n));
}
}
 
// This code is contributed
// by ChitraNayal

PHP

<?php
// PHP program to find OR
// of all the sub-arrays
 
// function to return OR
// of sub-arrays
function O_R($a, $n)
{
    $ans = $a[0];
    for ($i = 1; $i < $n; ++$i)
        $ans |= $a[$i];
 
    return $ans;
}
 
// Driver Code
$a = array( 1, 4, 6 );
$n = count($a);
 
// print OR of all subarrays
echo O_R($a, $n);
 
// This code is contributed
// by inder_verma
?>

Javascript

<script>
 
// Javascript program to find OR of all the sub-arrays
 
// function to return OR of sub-arrays
function OR(a, n)
{
    var ans = a[0];
    for (var i = 1; i < n; ++i)
         ans |= a[i];   
 
    return ans;
}
 
var a = [ 1, 4, 6 ];
var n = a.length;
 
    // print OR of all subarrays
 document.write(OR(a, n));
 
// This code is contributed by SoumikMondal
 
</script>

Output:

Time Complexity: O(N)

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