Coding problem · Easy · Basics

Leap Year

Easy Algorithms & data structures

Return true if year is a leap year:

Input: year = 2024
Output: true

Input: year = 1900
Output: false

Input: year = 2000
Output: true

Hints

Hint 1

Combine the rules with && and ||: divisible by 4 and (not by 100, or by 400).

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Time O(1) · Space O(1)

year % 4 == 0 && (year % 100 != 0 || year % 400 == 0). 1900 fails the century rule; 2000 is rescued by the 400 rule.

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

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

def is_leap_year(year: int) -> bool:
    return year % 4 == 0 and (year % 100 != 0 or year % 400 == 0)

<?php

function isLeapYear(int $year): bool
{
    return $year % 4 === 0 && ($year % 100 !== 0 || $year % 400 === 0);
}

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

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