Java Program to Check Even or Odd Number

In Java, you can check if a number is even or odd by using the modulus operator % like this: if (number % 2 == 0) { System.out.println("Even"); } else { System.out.println("Odd"); }.

Examples

Input4

Output4 is Even

Input7

Output7 is Odd

Input0

Output0 is Even

How to Think About It

To check if a number is even or odd, divide it by 2 and look at the remainder. If the remainder is 0, the number is even because it divides evenly by 2. If the remainder is not 0, the number is odd because it leaves a remainder when divided by 2.

Algorithm

1

Get the input number from the user.

2

Calculate the remainder when the number is divided by 2 using the modulus operator.

3

If the remainder is 0, print that the number is even.

4

Otherwise, print that the number is odd.

Code

java

import java.util.Scanner;

public class EvenOddChecker {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        System.out.print("Enter a number: ");
        int number = scanner.nextInt();
        if (number % 2 == 0) {
            System.out.println(number + " is Even");
        } else {
            System.out.println(number + " is Odd");
        }
        scanner.close();
    }
}

Output

Enter a number: 7
7 is Odd

Dry Run

Let's trace the input 7 through the code

1

Input number

User enters 7, so number = 7

2

Calculate remainder

7 % 2 = 1 (since 7 divided by 2 leaves remainder 1)

3

Check remainder

Remainder is 1, which is not 0, so number is odd

4

Print result

Prints '7 is Odd'

Step	Operation	Value
1	Input number	7
2	7 % 2	1
3	Check if remainder == 0	False
4	Print output	7 is Odd

Why This Works

Step 1: Using modulus operator

The % operator gives the remainder of division, which helps us know if a number divides evenly by 2.

Step 2: Even number condition

If the remainder is 0, it means the number is divisible by 2 and is even.

Step 3: Odd number condition

If the remainder is not 0, the number is not divisible by 2 and is odd.

Alternative Approaches

Using bitwise AND operator

java

import java.util.Scanner;

public class EvenOddChecker {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        System.out.print("Enter a number: ");
        int number = scanner.nextInt();
        if ((number & 1) == 0) {
            System.out.println(number + " is Even");
        } else {
            System.out.println(number + " is Odd");
        }
        scanner.close();
    }
}

This method uses bitwise AND with 1 to check the last bit; it's often faster but less intuitive for beginners.

Using ternary operator

java

import java.util.Scanner;

public class EvenOddChecker {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        System.out.print("Enter a number: ");
        int number = scanner.nextInt();
        String result = (number % 2 == 0) ? "Even" : "Odd";
        System.out.println(number + " is " + result);
        scanner.close();
    }
}

This approach uses a compact ternary operator for cleaner code but does the same check.

Complexity: O(1) time, O(1) space

Time Complexity

The check uses a single modulus operation which takes constant time regardless of input size.

Space Complexity

Only a few variables are used, so the space needed is constant.

Which Approach is Fastest?

Both modulus and bitwise AND methods run in constant time; bitwise AND can be slightly faster but is less readable.

Approach	Time	Space	Best For
Modulus operator (%)	O(1)	O(1)	Readability and simplicity
Bitwise AND (&)	O(1)	O(1)	Performance in low-level or large-scale systems
Ternary operator	O(1)	O(1)	Compact code style