# Java Program to Check if a Given Integer is Odd or Even

A number that is divisible by 2 and generates a remainder of 0 is called an even number. All the numbers ending with 0, 2, 4, 6, and 8 are even numbers. On the other hand, number that is not divisible by 2 and generates a remainder of 1 is called an odd number. All the numbers ending with 1, 3, 5,7, and 9 are odd numbers. Do refer to the below illustration to get what is supposed to be conveyed out basics here via generic Illustration for any random integer, check whether it is even or odd.

```Input : 13
Output: ODD```
```Input : 24
Output: EVEN```

Methods:

There are various ways to check whether the given number is odd or even. Some of them are as follows starting from the brute force approach ending up at the most optimal approach.

1. Using Brute Force- Naive Approach
2. Using bitwise operators
• Using Bitwise OR
• Using Bitwise AND
• Using Bitwise XOR
3. By Checking the Least Significant Bit

Method 1: Brute Force Naive Approach

It is to check the remainder after dividing by 2. Numbers that are divisible by 2 are even else odd.

Example

## Java

 `// Java Program to Check if Given Integer is Odd or Even` `// Using Brute Forcew Approach`   `// Importing required classes` `import` `java.io.*;` `import` `java.util.Scanner;`   `// Main class` `class` `GFG {`   `    ``// Main Driver Method` `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``// Declaring and initializing integer variable` `        ``int` `num = ``10``;`   `        ``// Checking if number is even or odd number` `        ``// via remainder` `        ``if` `(num % ``2` `== ``0``) {`   `            ``// If remainder is zero then this number is even` `            ``System.out.println(``"Entered Number is Even"``);` `        ``}`   `        ``else` `{`   `            ``// If remainder is not zero then this number is` `            ``// odd` `            ``System.out.println(``"Entered Number is Odd"``);` `        ``}` `    ``}` `}`

Output

`Entered Number is Even`

Time Complexity: O(1)

Auxiliary Space: O(1)

Now let us dwell on optimal approaches as follows below as follows with help of bitwise operators

Method 2: Using bitwise operators

• Bitwise OR
• Bitwise AND or Bitwise XOR

2-A: Using Bitwise OR

Bitwise OR operation of the even number by 1 increment the value of the number by 1 otherwise it remains unchanged.

Illustration: Bitwise OR

```    Number = 12              1  1  0  0    - Representation of 12 in Binary Format
Bitwise OR   0  0  0  1    - Representation of  1 in Binary Format

1  1  0  1    - Representation of 13 in Binary Format
Result- Number was even so bitwise Or by 1 increment the value by 1```
```    Number = 15            1  1  1  1    - Representation of 15 in Binary Format
Bitwise OR  0  0  0  1    - Representation of  1 in Binary Format

1  1  1  1    - Representation of 15 in Binary Format
Result- Number was odd so bitwise Or by 1 doesn't increment the value by 1```

Example

## Java

 `// Java Program to Check if Given Integer is Odd or Even` `// Using Bitwise OR`   `// Importing required classes` `import` `java.util.*;`   `// Main class` `public` `class` `GFG {`   `    ``// Main driver method` `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``// Declaring and initializing integer variable` `        ``// to be checked` `        ``int` `n = ``100``;`   `        ``// Condition check` `        ``// if n|1 if greater than n then this number is even` `        ``if` `((n | ``1``) > n) {`   `            ``// Print statement` `            ``System.out.println(``"Number is Even"``);` `        ``}` `        ``else` `{`   `            ``// Print statement` `            ``System.out.println(``"Number is Odd"``);` `        ``}` `    ``}` `}`

Output

`Number is Even`

Time Complexity: O(1)

Auxiliary Space: O(1)

2-B: Using Bitwise AND

Bitwise AND operation of the odd number by 1 will be 1 because the last bit will be already set otherwise it will give 0.

Illustration: Bitwise AND

```    Number = 5              0  1  0  1    - Representation of  5 in Binary Format
Bitwise AND  0  0  0  1    - Representation of  1 in Binary Format

0  0  0  1    - Representation of  1 in Binary Format
Result- Number was odd so bitwise And by 1 is 1```
```        Number = 8            1  0  0  0    - Representation of  8 in Binary Format
Bitwise AND  0  0  0  1    - Representation of  1 in Binary Format

0  0  0  0    - Representation of  0 in Binary Format
Result- Number was even so bitwise And by 1 is 0```

Example

## Java

 `// Java Program to Check if Given Integer is Odd or Even` `// Using Bitwise AND`   `// Importing required classes` `import` `java.util.*;`   `// Main class` `public` `class` `GFG {`   `    ``// Main driver method` `    ``public` `static` `void` `main(String[] args)` `    ``{`   `        ``// Declare and initializing integer variable` `        ``int` `n = ``91``;`   `        ``// Condition Check` `        ``// Bitwise AND of any odd number by 1 gives 1` `        ``if` `((n & ``1``) == ``1``) {`   `            ``// Print statement` `            ``System.out.println(``"Number is Odd"``);` `        ``}` `        ``else` `{`   `            ``// Print statement` `            ``System.out.println(``"Number is Even"``);` `        ``}` `    ``}` `}`

Output

`Number is Odd`

Time Complexity: O(1)

Auxiliary Space: O(1)

2-C: Using Bitwise XOR

Bitwise XOR operation of the even number by 1 increment the value of the number by 1 otherwise it decrements the value of the number by 1 if the value is odd. It is the most optimal approach.

Illustration: Bitwise XOR

```    Number = 5              0  1  0  1    - Representation of  5 in Binary Format
Bitwise XOR  0  0  0  1    - Representation of  1 in Binary Format

0  1  0  0    - Representation of  4 in Binary Format
Result- Number was odd so bitwise And by 1 decrement the value```
```    Number = 8            1  0  0  0    - Representation of  8 in Binary Format
Bitwise XOR  0  0  0  1    - Representation of  1 in Binary Format

1  0  0  1    - Representation of  9 in Binary Format
Result- Number was even so bitwise And by 1 increment the value```

Example

## Java

 `// Java Program to Check if Given Integer is Odd or Even` `// Using Bitwise XOR`   `// Importing required classes` `import` `java.util.*;`   `// Main class` `public` `class` `GFG {`   `    ``// Main driver method` `    ``public` `static` `void` `main(String[] args)` `    ``{`   `        ``// Declare and initializing integer variable` `        ``int` `num = ``99``;`   `        ``// Condition Check` `        ``// if number^1 increments by 1 then its even number,` `        ``// else odd` `        ``if` `((num ^ ``1``) == num + ``1``) {`   `            ``// Print statement` `            ``System.out.println(``"Number is Even"``);` `        ``}` `        ``else` `{`   `            ``// Print statement` `            ``System.out.println(``"Number is Odd"``);` `        ``}` `    ``}` `}`

Output

`Number is Odd`

Time Complexity: O(1)

Auxiliary Space: O(1)

Method 3: Checking the LSB of the Number

The LSB(Least Significant Bit) of an even number is always 0 and that of an odd number is always 1.

Example

## Java

 `// Java Program to Check if Given Integer is Odd or Even` `// by checking the LSB of the Number`   `// Importing required classes` `import` `java.util.*;`   `// Main class` `// TestEvenOddByCheckingLSB` `public` `class` `GFG {`   `    ``// Method 1` `    ``// To test number is even or odd` `    ``public` `static` `String testOddEvenByCheckingLSB(``int` `a)` `    ``{`   `        ``if` `(a != ``0``) {` `            ``if` `(Integer.toBinaryString(a).endsWith(``"0"``)) {` `                ``return` `"Even"``;` `            ``}` `            ``else` `{` `                ``return` `"Odd"``;` `            ``}` `        ``}`   `        ``// Here we will land if` `        ``// it does not ends with 0` `        ``else` `{` `            ``return` `"Zero"``;` `        ``}` `    ``}`   `    ``// Method 2` `    ``// Main driver method` `    ``public` `static` `void` `main(String[] args)` `    ``{`   `        ``// Iterationg over using for loop` `        ``for` `(``int` `i = ``0``; i <= ``10``; i++) {`   `            ``// Calling the function and printing` `            ``// corresponding number is even or odd` `            ``System.out.println(` `                ``i + ``" : "` `+ testOddEvenByCheckingLSB(i));` `        ``}` `    ``}` `}`

Output

```0 : Zero
1 : Odd
2 : Even
3 : Odd
4 : Even
5 : Odd
6 : Even
7 : Odd
8 : Even
9 : Odd
10 : Even```

Time Complexity: O(1)

Auxiliary Space: O(1)

Whether you're preparing for your first job interview or aiming to upskill in this ever-evolving tech landscape, GeeksforGeeks Courses are your key to success. We provide top-quality content at affordable prices, all geared towards accelerating your growth in a time-bound manner. Join the millions we've already empowered, and we're here to do the same for you. Don't miss out - check it out now!

Previous
Next