---
Title: Java Program to Check Leap Year - Scaler Topics
description: In this article, Scaler Topics will teach you how to identify a leap year and showcase a Java program to check whether the given year is a leap year.
author: Abhishek Shree
Category: Java

---
:::section{.abstract}
<!-- ## Introduction -->

Leap years are those years with 366 days instead of 365. A leap year must satisfy any one of the following two constraints:

* It should be divisible by 400
* It should be divisible by four and not by 100


:::
::: section{methods}
## Methods for Leap Year Program in Java
Java provides specific methods to evaluate leap year, which are mentioned below:

1. Using the Ternary Operator for Finding Leap Year
2. Using command-line arguments to check for Leap Year in Java
3. Using Scanner class to check for Leap Year in Java
4. Using if-else statements for Finding Leap Year in Java
5. Using in-built isLeap() method to check for Leap Year in Java
:::

:::section{.main}
* ## Using the Ternary Operator for Finding Leap Year Program in Java

The basic idea of a `ternary operator` in Java is to reduce the `if-else statements`, as those can get cumbersome, as we saw in the earlier method.
A general structure for the ternary operator is shown in the image below.


A program to find whether a year is a leap or not using ternary operators in Java will need `three nested ternary` operators, one for each of the conditions mentioned in our flowchart.

**Code:**

public class LeapYear {

  public static void main(String[] args) {
    // The desired year to check.
    int year = 1998;

    String result;

    result =
      (
        (year % 4 == 0 && year % 100 != 0)
          ? "is a leap year."
          : (year % 400 == 0) ? "is a leap year." : "is not a leap year."
      );

    // Append the year and format the string
    System.out.println(year + " " + result);
  }
}

**Output:**

1998 is not a leap year.

### Complexity
Time Complexity- O(1)
Space Complexity- O(1)

:::

:::section{.main}
* ## Using command-line arguments to check for Leap Year in Java

`Command-line` arguments are stored as strings in the String array passed to the main method in Java classes. We can use this feature to take input directly while running our program after compilation.

1. Here, we take the user input directly from the command line when the Java compiled code is executed.
2. We read arguments from `String[] args` in the `main` method using array syntax and convert them into integers by using the `Integer` class method `Integer.parseInt()` which converts other data types to integers.
3. We place the input in the `checkLeapYear(int year)` method and display whether the given year is a `leap year or not`.

**Code:**

public class LeapYear {

  // Method to check leap year
  public static void checkLeapYear(int year) {
    if (year % 400 == 0) {
      System.out.println(year + " is a leap year.");
    } else if (year % 100 == 0) {
      System.out.println(year + " is not a leap year.");
    } else if (year % 4 == 0) {
      System.out.println(year + " is a leap year.");
    } else {
      System.out.println(year + " is not a leap year.");
    }
  }

  public static void main(String[] args) {
    if (args.length > 0) {
      // convert string into integer
      int year = Integer.parseInt(args[0]);
      checkLeapYear(year);
    } else {
      System.out.println("No arguments provided.");
    }
  }
}

How to run:
Store the code in a file named `LeapYear.java`, compile it using:

javac LeapYear.java

Execute the class file generated using the following command:

java LeapYear 1998

**Output:**

1998 is not a leap year.

### Complexity
Time Complexity- O(1)
Space Complexity- O(1)

`Command-line` arguments open a completely new domain for taking inputs in java programs and can be helpful at times.

* ### Using Scanner class to check for Leap Year in Java
We can access the `Scanner` class to take inputs from the command line in Java. The process is as follows:

1. We import the `Scanner` class from the **Java.util** package.
2. In the `main` method, we create a `Scanner` object (named `s`).
3. Then we scan an integer using `s.nextInt()` method and store it in an integer variable.
4. We supply the value to the `checkLeapYear(int year)` method and display whether the given year is a leap year or not.

**Code:**

import java.util.Scanner; // Import the Scanner class

public class LeapYear {

  // Method to check leap year
  public static void checkLeapYear(int year) {
    if (year % 400 == 0) {
      System.out.println(year + " is a leap year.");
    } else if (year % 100 == 0) {
      System.out.println(year + " is not a leap year.");
    } else if (year % 4 == 0) {
      System.out.println(year + " is a leap year.");
    } else {
      System.out.println(year + " is not a leap year.");
    }
  }

  public static void main(String[] args) {
    Scanner s = new Scanner(System.in); // Create a Scanner object
    System.out.println("Enter a year to check:");

    int year = s.nextInt(); // Read user input from command line

    checkLeapYear(year);
  }
}

**Output:**

Enter a year to check:     
1998                       
1998 is not a leap year.

### Complexity
Time Complexity- O(1)
Space Complexity- O(1)
:::
:::section{.main}
* ## Using if-else statements for Finding Leap Program Year in Java
The procedure is as follows:

1. Let the year be `y`.
2. We start our `if-else` block and check if the year is `divisible by 400`, if not, if true, the year is a leap year, and we print the same, Otherwise we move to the else if block.
3. If the first else if block, we check if the year is `divisible by 100`. if true, the year is not a leap year, and we print the result; otherwise, we move to the next if block.
4. In the last else if block, we check if the year is `divisible by 4`, and if that turns out to be true the year is a leap year and we display the result, if that condition fails too, we conclude that the year is `not a leap year` in our final else block.

### Code

public class LeapYear {

  public static void main(String[] args) {
    // The desired year to check.
    int year = 1998;

    // Implementing our algorithm.
    if (year % 400 == 0) {
      System.out.println(year + " is a leap year.");
    } 
    else if (year % 100 == 0) {
      System.out.println(year + " is not a leap year.");
    } 
    else if (year % 4 == 0) {
      System.out.println(year + " is a leap year.");
    } 
    else {
      System.out.println(year + " is not a leap year.");
    }
  }
}

### Output

1998 is not a leap year.

### Complexity
Time Complexity- O(1)
Space Complexity- O(1)
<!--
### Explanation
* Here, we are checking all the conditions one by one to get the desired result.
* Altering the order of statements will `deviate the code` from the `flowchart`, leading to undesired outputs.
* We can also use `logical` operators to reduce the number of `if-else` statements in the first snippet.

public class LeapYear {

  public static void main(String[] args) {
    int year = 1998;
    if ((year % 400 == 0) || ((year % 4 == 0) && (year % 100 != 0))) {
      System.out.println(year + " is a leap year.");
    } else {
      System.out.println(year + " is not a leap year.");
    }
  }
}

**Output:**

1998 is not a leap year.

**Note:** We should generally avoid complex combinations of `logical operators` as it reduces the readability of our code and makes it more `prone to bugs`.
-->
:::
:::section{.main}
* ## Using in-built isLeap() method to check for Leap Year in Java

Java provides an in-built method isLeap() in the Year class to identify if an input year is leap or not. Let us see its implementation:

**Declaration:**

public boolean isLeap()

**Code:**

// isLeap() method in Java illustration

import java.time.*;
import java.util.*;

public class CheckLeapYear {

  public static void main(String[] args) {
    // Create the Year object: year
    Year year = Year.of(2020);

    // Print the result of isLeap method
    System.out.println(year.isLeap());
  }
}

**Output:**

true

### Complexity
Time Complexity- O(1)
Space Complexity- O(1)
:::
:::section{.summary}
## Conclusion
* Leap years are those years that are either divisible by 400 or they are divisible by four and not by 100.
* We can find leap years using simple if-else blocks.
* We can use ternary operators to present the if-else logic in an elegant manner.
* To make the code modular, the whole if-else logic can be shifted to a method that decides and prints whether the input year is leap or not.
* Apart from implementing it from scratch, Java also has a `Year` class with inbuilt methods like `isLeap()` to ease some of our work.

:::