Ruby Program to Check Leap Year with Output and Explanation

In Ruby, you can check a leap year using if (year % 400 == 0) || (year % 4 == 0 && year % 100 != 0) to print whether the year is leap or not.

Examples

Input2000

Output2000 is a leap year

Input1900

Output1900 is not a leap year

Input2024

Output2024 is a leap year

How to Think About It

To check if a year is leap, first see if it divides evenly by 400. If yes, it is leap. If not, check if it divides evenly by 4 but not by 100. If yes, it is leap. Otherwise, it is not leap.

Algorithm

1

Get the input year from the user

2

Check if the year is divisible by 400; if yes, it is a leap year

3

Else, check if the year is divisible by 4 but not by 100; if yes, it is a leap year

4

Otherwise, it is not a leap year

5

Print the result

Code

ruby

puts "Enter a year:"
year = gets.to_i
if (year % 400).zero? || ((year % 4).zero? && !(year % 100).zero?)
  puts "#{year} is a leap year"
else
  puts "#{year} is not a leap year"
end

Output

Enter a year:
2024
2024 is a leap year

Dry Run

Let's trace the year 2024 through the code

1

Input year

year = 2024

2

Check divisibility by 400

2024 % 400 = 24 (not zero)

3

Check divisibility by 4 and not by 100

2024 % 4 = 0 (zero), 2024 % 100 = 24 (not zero)

4

Determine leap year

Condition true, so 2024 is a leap year

Year	year % 400	year % 4	year % 100	Leap Year?
2024	24	0	24	Yes

Why This Works

Step 1: Divisible by 400

A year divisible by 400 is always a leap year, so we check year % 400 == 0 first.

Step 2: Divisible by 4 but not 100

If not divisible by 400, a year divisible by 4 but not by 100 is a leap year, checked by year % 4 == 0 && year % 100 != 0.

Step 3: Otherwise not leap

If neither condition is true, the year is not a leap year.

Alternative Approaches

Using a method to return boolean

ruby

def leap_year?(year)
  (year % 400).zero? || ((year % 4).zero? && !(year % 100).zero?)
end

puts "Enter a year:"
year = gets.to_i
if leap_year?(year)
  puts "#{year} is a leap year"
else
  puts "#{year} is not a leap year"
end

This approach separates logic into a method for reuse and clarity.

Using case statement

ruby

puts "Enter a year:"
year = gets.to_i
case
when (year % 400).zero?
  puts "#{year} is a leap year"
when (year % 4).zero? && !(year % 100).zero?
  puts "#{year} is a leap year"
else
  puts "#{year} is not a leap year"
end

Using case without argument can make conditions clearer but is slightly longer.

Complexity: O(1) time, O(1) space

Time Complexity

The program performs a fixed number of arithmetic operations and checks, so it runs in constant time.

Space Complexity

Only a few variables are used, so the space used is constant.

Which Approach is Fastest?

All approaches run in constant time; using a method improves readability but does not affect speed.

Approach	Time	Space	Best For
Inline if-else	O(1)	O(1)	Simple scripts
Method returning boolean	O(1)	O(1)	Reusable code
Case statement	O(1)	O(1)	Clear condition grouping