🚀Day 23/180 To check if the number is even or odd using Bitmanipulation...

To check if the number is even or odd using Bitmanipulation...

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// Now using the concept of BitManipulation we can find if the number is even or odd
// using the & operator 
// So, the concept here is ==> Even & ==> then 0 at last 
// Odd & ==> then 1 at last ...

import java.util.*;

public class BitManipulation {
    public static String evenOdd(int n){
        // I guesss 
        int bitMask =1;
            if ((n & bitMask)==1){
                return "Odd";
                }
            else {
                return "Even";
                }
        }

    public static void main (String args []){
            Scanner scanner = new Scanner(System.in);
            System.out.println ("Enter the number for which you want to check if it is even or odd");
        int num = scanner.nextInt();
        System.out.println(evenOdd(num));
        }
    } 
/*
PS C:\Users\HP\OneDrive\Coding\Java\Practice01> javac BitManipulation.java
PS C:\Users\HP\OneDrive\Coding\Java\Practice01> java BitManipulation     
Enter the number for which you want to check if it is even or odd
45
Odd

Let's go through a dry run of the evenOdd method to understand how it determines if a number is even or odd using bit manipulation.

Here's the method:

public static String evenOdd(int n) {
    int bitMask = 1;
    if ((n & bitMask) == 1) {
        return "Odd";
    } else {
        return "Even";
    }
}

Explanation

• The & operator is a bitwise AND operator.

• A bitmask is used to isolate a specific bit of a number. In this case, bitMask = 1 which in binary is 0001.

• When n is ANDed with bitMask, it checks the least significant bit (the rightmost bit) of n.

  • If the least significant bit is 1, the number is odd.

  • If the least significant bit is 0, the number is even.

Dry Run

Let's consider the example input 45.

1. Convert 45 to binary:

   • Decimal 45 is 101101 in binary.

2. Apply the bitmask:

   • Bitmask 0001

   • 45 in binary 101101

   • Perform the bitwise AND operation:

       101101
     

& 000001

000001

    - The result is `1`, which means the number is odd.

3. Method logic:

   • The condition (n & bitMask) == 1 evaluates to true because 45 & 1 is 1.

   • The method returns "Odd".

Testing the Code

When you run the code and input 45, it correctly prints Odd.

Here's the complete program for reference:

import java.util.*;

public class BitManipulation {
    public static String evenOdd(int n) {
        int bitMask = 1;
        if ((n & bitMask) == 1) {
            return "Odd";
        } else {
            return "Even";
        }
    }

    public static void main(String args[]) {
        Scanner scanner = new Scanner(System.in);
        System.out.println("Enter the number for which you want to check if it is even or odd:");
        int num = scanner.nextInt();
        System.out.println(evenOdd(num));
    }
}

This code correctly uses bit manipulation to determine if a number is even or odd.