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# lainjiang's solution

## to Leap in the JavaScript Track

Published at Jul 13 2018 · 6 comments
Instructions
Test suite
Solution

#### Note:

This solution was written on an old version of Exercism. The tests below might not correspond to the solution code, and the exercise may have changed since this code was written.

Given a year, report if it is a leap year.

The tricky thing here is that a leap year in the Gregorian calendar occurs:

``````on every year that is evenly divisible by 4
except every year that is evenly divisible by 100
unless the year is also evenly divisible by 400
``````

For example, 1997 is not a leap year, but 1996 is. 1900 is not a leap year, but 2000 is.

If your language provides a method in the standard library that does this look-up, pretend it doesn't exist and implement it yourself.

## Notes

Though our exercise adopts some very simple rules, there is more to learn!

For a delightful, four minute explanation of the whole leap year phenomenon, go watch this youtube video.

## Setup

Go through the setup instructions for ECMAScript to install the necessary dependencies:

http://exercism.io/languages/ecmascript

## Requirements

Install assignment dependencies:

``````\$ npm install
``````

## Making the test suite pass

Execute the tests with:

``````\$ npm test
``````

In the test suites all tests but the first have been skipped.

Once you get a test passing, you can enable the next one by changing `xtest` to `test`.

## Source

JavaRanch Cattle Drive, exercise 3 http://www.javaranch.com/leap.jsp

## Submitting Incomplete Solutions

It's possible to submit an incomplete solution so you can see how others have completed the exercise.

### leap.spec.js

``````import Year from './leap';

describe('A leap year', () => {
test('year not divisible by 4: common year', () => {
const year = new Year(2015);
expect(year.isLeap()).toBeFalsy();
});

xtest('year divisible by 4, not divisible by 100: leap year', () => {
const year = new Year(2016);
expect(year.isLeap()).toBeTruthy();
});

xtest('year divisible by 100, not divisible by 400: common year', () => {
const year = new Year(2100);
expect(year.isLeap()).toBeFalsy();
});

xtest('year divisible by 400: leap year', () => {
const year = new Year(2000);
expect(year.isLeap()).toBeTruthy();
});
});``````
``````var isLeapYear = function(year) {
return year % 100 === 0 ? year % 400 === 0 : year % 4 === 0;
};

export default isLeapYear;``````

## Community comments

Find this solution interesting? Ask the author a question to learn more.

Really nice!

I've seen solutions to this exercise using logical operators to compute three conditions, but this ternary operator surprised me. I like it!

Agreed, very clever. Most solutions either funnel in (400 -> 100 -> 4) or expand out (4 and !100 or 400). Interesting to see (1) conditional operator, (2) using the result of 100 to decide whether to test 400 or 4, and (3) no use of not or false.

I've seen many solutions (on this site and another), but never this approach.

Indeed, awesome solution. Programming approach for thought

Very smart indeed. Hands up!

Very clean, in comparison to mine where I just have `if`'s and `return`'s :(

Very very Nice!

### What can you learn from this solution?

A huge amount can be learned from reading other peopleโs code. This is why we wanted to give exercism users the option of making their solutions public.

Here are some questions to help you reflect on this solution and learn the most from it.

• What compromises have been made?
• Are there new concepts here that you could read more about to improve your understanding?