introduction.md

Introduction

There are various idiomatic approaches to solve Leap. You can use a chain of boolean expressions to test the conditions. Or you can use a ternary operator.

General guidance

The key to solving Leap is to know if the year is evenly divisible by 4, 100 and 400. For determining that, you will use the remainder operator.

Approach: Chain of Boolean expressions

boolean isLeapYear(int year) {
    return year % 4 == 0 && (year % 100 != 0 || year % 400 == 0);
}

For more information, check the Boolean chain approach.

Approach: Ternary operator of Boolean expressions

boolean isLeapYear(int year) {
    return year % 100 == 0 ? year % 400 == 0 : year % 4 == 0;
}

For more information, check the Ternary operator approach.

Other approaches

Besides the aforementioned, idiomatic approaches, you could also approach the exercise as follows:

plusDays() approach

Add a day to February 28th for the year and see if the new day is the 29th. For more information, see the plusDays() approach.

Built-in method approach

Use the built-in method for the Year. For more information, see the isLeap() approach.

Which approach to use?

• The chain of boolean expressions is most efficient, as it proceeds from the most likely to least likely conditions. It has a maximum of three checks.
• The ternary operator has a maximum of only two checks, but it starts from a less likely condition.
• Using plusDays() or using the built-in isLeap() method may be considered "cheats" for the exercise, but isLeap() would be the idiomatic way to check if a year is a leap year in Java.