Bitwise OR( | ) of all even number from 1 to N

Given a number N, the task is to find the bitwise OR( | ) of all even numbers from 1 to N.

Examples:

Input: 2 
Output: 2

Input: 10 
Output: 14 
Explanation: 2 | 4 | 6 | 8 | 10 = 14 

Naive Approach:

• Initialize the result as 2.
• Iterate the loop from 4 to n (for all even number) and update result by finding bitwise or ( | ).

Below is the implementation of the approach: 

C++

// C++ implementation of the above approach
#include <iostream>
using namespace std;

// Function to return the bitwise OR
// of all the even numbers upto N
int bitwiseOrTillN(int n)
{
    // Initialize result as 2
    int result = 2;

    for (int i = 4; i <= n; i = i + 2) {
        result = result | i;
    }
    return result;
}

// Driver code
int main()
{
    int n = 10;
    cout << bitwiseOrTillN(n);
    return 0;
}

Java

// Java implementation of the above approach
class GFG
{

    // Function to return the bitwise OR
    // of all the even numbers upto N
    static int bitwiseOrTillN(int n)
    {
        // Initialize result as 2
        int result = 2;

        for (int i = 4; i <= n; i = i + 2)
        {
            result = result | i;
        }
        return result;
    }

    // Driver code
    static public void main (String args[])
    {
        int n = 10;
        System.out.println(bitwiseOrTillN(n));
    }
}

// This code is contributed by AnkitRai01

Python3

# Python 3 implementation of the above approach

# Function to return the bitwise OR
# of all the even numbers upto N
def bitwiseOrTillN ( n ):

    # Initialize result as 2
    result = 2;

    for i in range(4, n + 1, 2) :
        result = result | i

    return result

# Driver code
n = 10;
print(bitwiseOrTillN(n));

# This code is contributed by ANKITKUMAR34

C#

// C# implementation of the above approach
using System;

class GFG
{

    // Function to return the bitwise OR
    // of all the even numbers upto N
    static int bitwiseOrTillN(int n)
    {
        // Initialize result as 2
        int result = 2;

        for (int i = 4; i <= n; i = i + 2)
        {
            result = result | i;
        }
        return result;
    }

    // Driver code
    static public void Main ()
    {
        int n = 10;
        Console.WriteLine(bitwiseOrTillN(n));
    }
}

// This code is contributed by AnkitRai01

Javascript

<script>

// Javascript implementation of the above approach

// Function to return the bitwise OR
// of all the even numbers upto N
function bitwiseOrTillN(n)
{

    // Initialize result as 2
    var result = 2;

    for(var i = 4; i <= n; i = i + 2)
    {
        result = result | i;
    }
    return result;
}

// Driver code
var n = 10;
document.write( bitwiseOrTillN(n));

// This code is contributed by noob2000

</script>

Output:

14

Time Complexity: O(n), where n is the given integer.
Auxiliary Space: O(1), no extra space is required, so it is a constant.

Efficient Approach: Compute the total number of bits in N. In bitwise OR, the rightmost bit will be 0 and all other bits will be 1. Therefore, return pow(2, total no. of bits)-2. It will give the equivalent value in decimal of bitwise OR.

Below is the implementation of the approach: 

C++

// C++ implementation of the above approach
#include <iostream>
#include <math.h>
using namespace std;

// Function to return the bitwise OR
// of all even numbers upto N
int bitwiseOrTillN(int n)
{
    // For value less than 2
    if (n < 2)
        return 0;

    // Count total number of bits in bitwise or
    // all bits will be set except last bit
    int bitCount = log2(n) + 1;

    // Compute 2 to the power bitCount and subtract 2
    return pow(2, bitCount) - 2;
}

// Driver code
int main()
{
    int n = 10;
    cout << bitwiseOrTillN(n);
    return 0;
}

Java

// Java implementation of the above approach
class GFG
{

    // Function to return the bitwise OR
    // of all even numbers upto N
    static int bitwiseOrTillN(int n)
    {
        // For value less than 2
        if (n < 2)
            return 0;

        // Count total number of bits in bitwise or
        // all bits will be set except last bit
        int bitCount = (int)(Math.log(n)/Math.log(2)) + 1;

        // Compute 2 to the power bitCount and subtract 2
        return (int)Math.pow(2, bitCount) - 2;
    }

    // Driver code
    public static void main (String[] args)
    {
        int n = 10;
        System.out.println(bitwiseOrTillN(n));
    }
}

// This code is contributed by AnkitRai01

Python3

# Python3 implementation of the above approach
from math import log2

# Function to return the bitwise OR
# of all even numbers upto N
def bitwiseOrTillN(n) :

    # For value less than 2
    if (n < 2) :
        return 0;

    # Count total number of bits in bitwise or
    # all bits will be set except last bit
    bitCount = int(log2(n)) + 1;

    # Compute 2 to the power bitCount and subtract 2
    return pow(2, bitCount) - 2;

# Driver code
if __name__ == "__main__" :

    n = 10;
    print(bitwiseOrTillN(n));

# This code is contributed by AnkitRai01

C#

// C# implementation of the above approach
using System;

class GFG
{

    // Function to return the bitwise OR
    // of all even numbers upto N
    static int bitwiseOrTillN(int n)
    {
        // For value less than 2
        if (n < 2)
            return 0;

        // Count total number of bits in bitwise or
        // all bits will be set except last bit
        int bitCount = (int)(Math.Log(n)/Math.Log(2)) + 1;

        // Compute 2 to the power bitCount and subtract 2
        return (int)Math.Pow(2, bitCount) - 2;
    }

    // Driver code
    public static void Main()
    {
        int n = 10;
        Console.WriteLine(bitwiseOrTillN(n));
    }
}

// This code is contributed by AnkitRai01

Javascript

<script>
// Javascript implementation of the approach

// Function to return the bitwise OR
// of all even numbers upto N
function bitwiseOrTillN(n)
{
    // For value less than 2
    if (n < 2)
        return 0;

    // Count total number of bits in bitwise or
    // all bits will be set except last bit
    var bitCount = parseInt(Math.log2(n) + 1);

    // Compute 2 to the power bitCount and subtract 2
    return Math.pow(2, bitCount) - 2;
}


// Driver code
var n = 10;
document.write(  bitwiseOrTillN(n));

//This code is contributed by SoumikMondal
</script>

Output:

14

Time Complexity: O(log(log n), where n is the given integer.
Auxiliary Space: O(1), no extra space is required, so it is a constant.