DPC

Posted on Mar 31

Daily JavaScript Challenge #JS-139: Determine Leap Year

#javascript #devchallenge #programming #webdev

Daily JavaScript Challenge: Determine Leap Year

Hey fellow developers! 👋 Welcome to today's JavaScript coding challenge. Let's keep those programming skills sharp!

The Challenge

Difficulty: Easy

Topic: Date Manipulation

Description

Create a function to determine if a given year is a leap year. A year is a leap year if it is divisible by 4. However, if the year is a century (divisible by 100), it must also be divisible by 400 to qualify as a leap year.

Ready to Begin?

https://www.dpcdev.com/

Fork this challenge
Write your solution
Test it against the provided test cases
Share your approach in the comments below!

Want to Learn More?

Check out the documentation about this topic here: https://en.wikipedia.org/wiki/Leap_year#Algorithm

Join the Discussion!

How did you approach this problem?
Did you find any interesting edge cases?
What was your biggest learning from this challenge?

Let's learn together! Drop your thoughts and questions in the comments below. 👇

This is part of our Daily JavaScript Challenge series. Follow me for daily programming challenges and let's grow together! 🚀

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Top comments (5)

Rahul Nainala • Apr 4

Const isLeapYear = (year) => {
if (year % 100 === 0) {
return year % 400 === 0;
}
return year % 4 === 0;
};

Vladas Saulis • Apr 1 • Edited

const isLeapYear = new Date(year, 1, 29).getMonth() === 1;

Massimo Artizzu • Apr 1

const isLeapYear = (year) => year % 4 === 0 &&
  (year % 100 !== 0 || year % 400 === 0);

Kumara • Apr 1

const isLeapYear = (year) => {
    if (year % 4 === 0) {
        if (year % 100 === 0) {
            return year % 400 === 0;
        }

        return true;
    }
    return false;
};

Antoine Raymond • Apr 3

You should avoid nested conditions at all costs.

Here is an alternative solution :

function isLeapYear(year) {
  if (year % 100 === 0) {
    return year % 400 === 0;
  }

  return year % 4 === 0;
}

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