# 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 number from 1 to N.

Examples:

Input: 2
Output: 2

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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 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

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 #include 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

Output:

14

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