Leap Year Program in Java

In this tutorial, we are going to discuss how to identify whether a given year is a leap year or not. But before proceeding further, we will be discussing the leap year.

Leap Year

A leap year, also known as a bissextile or intercalary year, contains 1 more additional day (a total of 366 days) as compare to other years. Usually, it comes at a gap of 4 years. 2008, 2012, 2016, etc., are some examples of leap years.

Rules to verify

A century year (a year ending with 00) is a leap year only, if it is divisible by 400. For example, 1700 is not a leap year as it is not divisible by 400. 2000 is a leap year as it is divisible by 400.
A non-century year is a leap year if the year is divisible by 4.

For example, 1996 is a leap year, and 1999 is not a leap year, as the former one is divisible by 4 and the latter one is not divisible by 4.

Implementation: Without Using Scanner Class

Observe the implementation of the above rules.

FileName: LeapYear.java

public class LeapYear 
{

// a method that checks whether the 
// year y is a leap year or not
boolean isLeapYr(int y)
{
// if condition that handles
// the century year
if((y % 100) == 0)
{
    if((y % 400) == 0)
    {
        return true;
    }
    else
    {
        return false;
    }

}

// if condition that handles the
// non century year
if((y % 4) == 0)
{
    return true;
}

return false;
}

// main method
public static void main(String argvs[])
{
// creating an object of the class LeapYear
LeapYear obj = new LeapYear();

// input 1
int year = 1996;

if(obj.isLeapYr(year))
{
    System.out.println("The year " + year + " is a leap year.");
}
else
{
    System.out.println("The year " + year + " is not a leap year.");
}

System.out.println();

// input 2
year = 1999;

if(obj.isLeapYr(year))
{
    System.out.println("The year " + year + " is a leap year.");
}
else
{
    System.out.println("The year " + year + " is not a leap year.");
}

System.out.println();

// input 3
year = 1700;

if(obj.isLeapYr(year))
{
    System.out.println("The year " + year + " is a leap year.");
}
else
{
    System.out.println("The year " + year + " is not a leap year.");
}

System.out.println();

// input 4
year = 2000;

if(obj.isLeapYr(year))
{
    System.out.println("The year " + year + " is a leap year.");
}
else
{
    System.out.println("The year " + year + " is not a leap year.");
}




}
}

Output:

The year 1996 is a leap year.

The year 1999 is not a leap year.

The year 1700 is not a leap year.

The year 2000 is a leap year.

Complexity Analysis: The time complexity of the program is O(1). The space complexity of the program is also O(1).

Implementation: With Using Scanner Class

Here, the user is provided the flexibility to enter the year of his/her choice, and then the program tests whether the entered number is a leap year or not. Also, instead of multiple if statements, we can use only one if do our job. The implementation of it is mentioned in the following program.

FileName: LeapYear1.java

// importing the Scanner Class
import java.util.Scanner;

public class LeapYear1 
{

// a method that checks whether the 
// year y is a leap year or not
boolean isLeapYr(int y)
{
// if condition that handles
// the century year as well as the non-century year
if (((y % 4 == 0) && (y % 100 != 0)) || (y % 400 == 0))
{
return true;
}

return false;
}

// main method
public static void main(String argvs[])
{
// creating an object of the class LeapYear1
LeapYear1 obj = new LeapYear1();

// we are taking input using the Scanner Class

Scanner scnObj = new Scanner(System.in);

System.out.println("Enter Year ");

// input 1
int year = scnObj.nextInt();

if(obj.isLeapYr(year))
{
    System.out.println("The year " + year + " is a leap year.");
}
else
{
    System.out.println("The year " + year + " is not a leap year.");
}

System.out.println();

System.out.println("Enter Year ");

// input 2
year = scnObj.nextInt();

if(obj.isLeapYr(year))
{
    System.out.println("The year " + year + " is a leap year.");
}
else
{
    System.out.println("The year " + year + " is not a leap year.");
}

System.out.println();

System.out.println("Enter Year ");

// input 3
year = scnObj.nextInt();

if(obj.isLeapYr(year))
{
    System.out.println("The year " + year + " is a leap year.");
}
else
{
    System.out.println("The year " + year + " is not a leap year.");
}

System.out.println();

System.out.println("Enter Year ");

// input 4
year = scnObj.nextInt();

if(obj.isLeapYr(year))
{
    System.out.println("The year " + year + " is a leap year.");
}
else
{
    System.out.println("The year " + year + " is not a leap year.");
}




}
}

Output:

Enter Year
1996
The year 1996 is a leap year.

Enter Year
1999
The year 1999 is not a leap year.

Enter Year
1700
The year 1700 is not a leap year.

Enter Year
2000
The year 2000 is a leap year.

Complexity Analysis: The time complexity of the program is O(1). The space complexity of the program is also O(1).

Implementation: Finding Leap Years Within a Range

We can also find leap years within a given range with the help of a for-loop. Observe the following program.

FileName: LeapYear2.java

public class LeapYear2
{

// a method that checks whether the 
// year y is a leap year or not
boolean isLeapYr(int y)
{
// if condition that handles
// the century year as well as the non-century year
if (((y % 4 == 0) && (y % 100 != 0)) || (y % 400 == 0))
{
return true;
}

return false;
}

// main method
public static void main(String argvs[])
{
// creating an object of the class LeapYear2
LeapYear2 obj = new LeapYear2();

// we will be finding the leap years
// within the range r1 and r2 (r1 and r2 inclusive)
int r1 = 2000;
int r2 = 2150;

System.out.println("Finding leap years within the range " + r1 + " and " + r2);
for(int i = r1; i <= r2; i++)
{
    if(obj.isLeapYr(i))
    {
        System.out.print(i + " ");
    }
}




}
}

Output:

Finding leap years within the range 2000 and 2150
2000 2004 2008 2012 2016 2020 2024 2028 2032 2036 2040 2044 2048 2052 2056 2060 2064 2068 2072 2076 2080 2084 2088 2092 2096 2104 2108 2112 2116 2120 2124 2128 2132 2136 2140 2144 2148

Complexity Analysis: The space and time complexity of the program is the same as the previous program.

Implementation: Using Strings for Finding the Range

We can also convert the years into strings and check for the leap years. See the following code.

FileName: LeapYear3.java

public class LeapYear3
{

// a method that checks whether the 
// year y is a leap year or not
boolean isLeapYr(String y)
{
int size = y.length();
int temp = 0;


// handling the case when years contain 4 or more than 4 digits
if(size >= 4)
{
    char ch1 = y.charAt(size - 1);
    char ch2 = y.charAt(size - 2);
    char ch3 = y.charAt(size - 3);
    char ch4 = y.charAt(size - 4);
    
    // for handling the century years
    if((ch1 == '0') && (ch2 == '0'))
    {
        
        String st1 = "" + ch4 + ch3;
        temp = Integer.parseInt(st1);
    }
    
    // for handling the non- century years
    else
    {
        String st1 = "" + ch2 + ch1;
        temp = Integer.parseInt(st1);
    }
}
// handling the case when years contain 3 digits
else if(size == 3)
{
    char ch1 = y.charAt(size - 1);
    char ch2 = y.charAt(size - 2);
    char ch3 = y.charAt(size - 3);
    
    // for handling the century years
    if((ch1 == '0') && (ch2 == '0'))
    {
        
        String st1 = "" + ch3;
        temp = Integer.parseInt(st1);
    }
    // for handling the non-century years
    else
    {
        String st1 = "" + ch2 + ch1;
        temp = Integer.parseInt(st1);
    }    
}

else if(size == 2)
{
char ch1 = y.charAt(size - 1);
char ch2 = y.charAt(size - 2);    
String st1 = "" + ch2 + ch1;
temp = Integer.parseInt(st1);
}

else
{
char ch1 = y.charAt(size - 1);
String st1 = "" + ch1;
temp = Integer.parseInt(st1);    
}

return ((temp % 4) == 0);
}

// main method
public static void main(String argvs[])
{
// creating an object of the class LeapYear3
LeapYear3 obj = new LeapYear3();

// we will be finding the leap years
// within the range r1 and r2 (r1 and r2 inclusive)
int r1 = 2000;
int r2 = 2150;

System.out.println("Finding leap years within the range " + r1 + " and " + r2);
for(int i = r1; i <= r2; i++)
{
    if(obj.isLeapYr(String.valueOf(i)))
    {
        System.out.print(i + " ");
    }
}




}
}

Output:

Finding leap years within the range 2000 and 2150
2000 2004 2008 2012 2016 2020 2024 2028 2032 2036 2040 2044 2048 2052 2056 2060 2064 2068 2072 2076 2080 2084 2088 2092 2096 2104 2108 2112 2116 2120 2124 2128 2132 2136 2140 2144 2148

Complexity Analysis: The program is converting the integer into String, that takes O(D) times. Also, program is storing that String using variable y, that also takes O(D) space. Therefore, the time as well the space complexity of the program is O(D), where D is the total number of digits present in the year for which we are checking leap year.