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

  • Last Updated : 05 May, 2021

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

Examples: 

Input:
Output: 2

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

Naive Approach: Initialize result as 2.Iterate 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

 

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

 


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