Find bitwise OR of all possible sub-arrays

Given an array A of size N where, . 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.

C++

#include <bits/stdc++.h>
using namespace std;

int OR(vector<int>& arr) {
    int n = arr.size();
    int res = 0;
    for (int i = 0; i < n; i++) {
        int curr = 0;
        for (int j = i; j < n; j++) {
            curr |= arr[j];
            res |= curr;
        }
    }
    return res;
}

int main() {
    vector<int> arr1 = {1, 4, 6};
    cout << OR(arr1) << endl;

    vector<int> arr2 = {10, 100, 1000};
    cout << OR(arr2) << endl;

    return 0;
}

Java

import java.util.*;

class Main {
    static int OR(List<Integer> arr) {
        int n = arr.size();
        int res = 0;
        for (int i = 0; i < n; i++) {
            int curr = 0;
            for (int j = i; j < n; j++) {
                curr |= arr.get(j);
                res |= curr;
            }
        }
        return res;
    }

    public static void main(String[] args) {
        List<Integer> arr1 = new ArrayList<>(Arrays.asList(1, 4, 6));
        System.out.println(OR(arr1));

        List<Integer> arr2 = new ArrayList<>(Arrays.asList(10, 100, 1000));
        System.out.println(OR(arr2));
    }
}

Python3

def OR(arr):
    n = len(arr)
    res = 0
    for i in range(n):
        curr = 0
        for j in range(i, n):
            curr |= arr[j]
            res |= curr
    return res

arr1 = [1,4,6]
print(OR(arr1))

arr2 = [10,100,1000]
print(OR(arr2))

C#

using System;
using System.Collections.Generic;

class Program {
    static int OR(List<int> arr) {
        int n = arr.Count;
        int res = 0;
        for (int i = 0; i < n; i++) {
            int curr = 0;
            for (int j = i; j < n; j++) {
                curr |= arr[j];
                res |= curr;
            }
        }
        return res;
    }

    static void Main(string[] args) {
        List<int> arr1 = new List<int>{1, 4, 6};
        Console.WriteLine(OR(arr1));

        List<int> arr2 = new List<int>{10, 100, 1000};
        Console.WriteLine(OR(arr2));
    }
}

Javascript

function OR(arr) {
const n = arr.length;
let res = 0;
for (let i = 0; i < n; i++) {
let curr = 0;
for (let j = i; j < n; j++) {
curr |= arr[j];
res |= curr;
}
}
return res;
}

console.log(OR([1, 4,6]));

Output

7
1006

Efficient Solution: Using the property that i:e it doesn’t matter how many times an element comes, it’s OR will be counted as one only. Thus our problem boils down to finding the OR of all the elements of the array.

Implementation:

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

7

Time Complexity: O(N)

Space Complexity: O(1)

Optimal Solution using Prefix Array for finding OR of all possible sub-arrays:-

Approach:-

• Create two arrays prefix_OR and suffix_OR, both of size N, to store the OR values of prefixes and suffixes respectively.
• Initialize the first element of prefix_OR with the first element of the input array, and the last element of suffix_OR with the last element of the input array.
• Traverse the input array from left to right, and fill prefix_OR using the formula prefix_OR[i] = prefix_OR[i-1] | arr[i].
• Traverse the input array from right to left, and fill suffix_OR using the formula suffix_OR[i] = suffix_OR[i+1] | arr[i].
• Initialize a variable res to 0, and traverse the input array. At each index i, calculate prefix_OR[i] | suffix_OR[i], and perform bitwise OR with res.
• Return the final value of res.

Here is the implementation of above approach:-

C++

#include <iostream>
#include <vector>
using namespace std;

int OR(vector<int>& arr) {
    int n = arr.size();
    vector<int> prefix_OR(n), suffix_OR(n);
    prefix_OR[0] = arr[0];
    suffix_OR[n - 1] = arr[n - 1];
    for (int i = 1; i < n; i++) {
        prefix_OR[i] = prefix_OR[i - 1] | arr[i];
    }
    for (int i = n - 2; i >= 0; i--) {
        suffix_OR[i] = suffix_OR[i + 1] | arr[i];
    }
    int res = 0;
    for (int i = 0; i < n; i++) {
        res |= prefix_OR[i] | suffix_OR[i];
    }
    return res;
}

int main() {
    vector<int> arr = {1, 4, 6};
    cout << OR(arr) << endl;

    arr = {10, 100, 1000};
    cout << OR(arr) << endl;

    return 0;
}

Java

import java.util.*;

public class Main {
    public static int OR(int[] arr) {
        int n = arr.length;
        int[] prefix_OR = new int[n];
        int[] suffix_OR = new int[n];
        prefix_OR[0] = arr[0];
        suffix_OR[n - 1] = arr[n - 1];
        for (int i = 1; i < n; i++) {
            prefix_OR[i] = prefix_OR[i - 1] | arr[i];
        }
        for (int i = n - 2; i >= 0; i--) {
            suffix_OR[i] = suffix_OR[i + 1] | arr[i];
        }
        int res = 0;
        for (int i = 0; i < n; i++) {
            res |= prefix_OR[i] | suffix_OR[i];
        }
        return res;
    }

    public static void main(String[] args) {
        int[] arr = {1, 4, 6};
        System.out.println(OR(arr));

        arr = new int[]{10, 100, 1000};
        System.out.println(OR(arr));
    }
}

Python3

def OR(arr):
    n = len(arr)
    prefix_OR = [0] * n
    suffix_OR = [0] * n
    prefix_OR[0] = arr[0]
    suffix_OR[n - 1] = arr[n - 1]
    for i in range(1, n):
        prefix_OR[i] = prefix_OR[i - 1] | arr[i]
    for i in range(n - 2, -1, -1):
        suffix_OR[i] = suffix_OR[i + 1] | arr[i]
    res = 0
    for i in range(n):
        res |= prefix_OR[i] | suffix_OR[i]
    return res


# Driver code
if __name__=='__main__':
  arr = [1, 4, 6]
  print(OR(arr))

  arr = [10, 100, 1000]
  print(OR(arr))

C#

using System;

public class Program
{
    public static int OR(int[] arr)
    {
        int n = arr.Length;
        int[] prefix_OR = new int[n];
        int[] suffix_OR = new int[n];
        prefix_OR[0] = arr[0];
        suffix_OR[n - 1] = arr[n - 1];
        for (int i = 1; i < n; i++)
        {
            prefix_OR[i] = prefix_OR[i - 1] | arr[i];
        }
        for (int i = n - 2; i >= 0; i--)
        {
            suffix_OR[i] = suffix_OR[i + 1] | arr[i];
        }
        int res = 0;
        for (int i = 0; i < n; i++)
        {
            res |= prefix_OR[i] | suffix_OR[i];
        }
        return res;
    }
    public static void Main()
    {
        int[] arr1 = { 1, 4, 6 };
        Console.WriteLine(OR(arr1));
        int[] arr2 = { 10, 100, 1000 };
        Console.WriteLine(OR(arr2));
    }
}

PHP

<?php
function OR($arr) {
    $n = count($arr);
    $prefix_OR = array();
    $suffix_OR = array();
    $prefix_OR[0] = $arr[0];
    $suffix_OR[$n - 1] = $arr[$n - 1];
    for ($i = 1; $i < $n; $i++) {
        $prefix_OR[$i] = $prefix_OR[$i - 1] | $arr[$i];
    }
    for ($i = $n - 2; $i >= 0; $i--) {
        $suffix_OR[$i] = $suffix_OR[$i + 1] | $arr[$i];
    }
    $res = 0;
    for ($i = 0; $i < $n; $i++) {
        $res |= $prefix_OR[$i] | $suffix_OR[$i];
    }
    return $res;
}

echo OR([1, 4, 6]);
?>

Javascript

function OR(arr) {
    const n = arr.length;
    const prefix_OR = new Array(n);
    const suffix_OR = new Array(n);
    prefix_OR[0] = arr[0];
    suffix_OR[n - 1] = arr[n - 1];
    for (let i = 1; i < n; i++) {
        prefix_OR[i] = prefix_OR[i - 1] | arr[i];
    }
    for (let i = n - 2; i >= 0; i--) {
        suffix_OR[i] = suffix_OR[i + 1] | arr[i];
    }
    let res = 0;
    for (let i = 0; i < n; i++) {
        res |= prefix_OR[i] | suffix_OR[i];
    }
    return res;
}

console.log(OR([1, 4, 6]));
console.log(OR([10, 100, 1000]));

Output

7
1006

Time Complexity:
The time complexity of this approach is O(N), where N is the size of the input array.

Space Complexity:
The space complexity of this approach is O(N), where N is the size of the input array, as we are using two extra arrays of size N to store the prefix and suffix OR values.