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?

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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

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Did you find any interesting edge cases?
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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)

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);

Rahul Nainala • Apr 4

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