Bitwise OR( | ) of all even number from 1 to N
Last Updated :
28 Dec, 2022
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 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
# 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# 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
<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>
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 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 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# 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
<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>
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.
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