L1 · FoundationsNumbers & loops~2 min · 6 tests

Check If a Year Is a Leap Year

Decide whether a year is a leap year with the full Gregorian rule: divisible by 4, except centuries, unless divisible by 400. Python conditions practice.

The problem

Return True if year is a leap year in the Gregorian calendar:

divisible by 4, except
years divisible by 100 are not, unless
they are also divisible by 400.

Examples

Example 1
Input
```
is_leap(2024)
```
Expected output
```
True
```
Example 2
Input
```
is_leap(1900)
```
Expected output
```
False
```
Example 3
Input
```
is_leap(2000)
```
Expected output
```
True
```

+ 3 hidden tests on Submit.

Edge cases to ask about

Century years (1900, 2100)
Multiples of 400

Hints

0/3

How an interviewer scores this

0/9

1. Understand the problemRestate it, ask about edge cases and input size.
2. Write a brute-force solutionCorrect first, clever later.
3. Get the correct outputautoAll visible examples pass.
4. Analyse time complexityautoCommit to a Big-O in the Complexity Lab.
5. Analyse space complexityautoExtra memory, not the input.
6. Identify the bottleneckautoMeasure your code — where does it bend?
7. OptimiseUsually: a set or dict instead of a scan.
8. Explain why it worksSay it out loud to Solvi.
9. Test the edge casesautoSubmit passes every hidden test.

Python 3.13 · is_leap

def is_leap(year):
    """True if year is a Gregorian leap year."""
    # Write your solution here
    pass

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Complexity Lab

What does this cost as n grows?

Interviewers score the analysis as much as the code. Commit to an answer first — then check it, and read why.

Time complexity of the optimal solution

Space complexity (extra memory)

Pick both to reveal the answer.

From brute force to optimal

The progression an interviewer wants to hear, one step at a time.

Approach	Time	Space	Idea
One boolean expression	O(1)	O(1)

Walkthrough of the optimal approach (try it yourself first)

year % 4 == 0 and (year % 100 != 0 or year % 400 == 0). 1900 is divisible by 4 and by 100 but not by 400, so it is not a leap year; 2000 is.

calendar.isleap exists — know it, but be able to write the rule.

Complexity: O(1) time, O(1) space. Three modulo checks, regardless of the year.

Reveal the reference solution

def is_leap(year):
    return year % 4 == 0 and (year % 100 != 0 or year % 400 == 0)

Follow-ups interviewers ask

Count the leap years between two years in O(1).

Frequently asked interview questions

Core interview concepts, complexities, and follow-ups scored by hiring teams.

What is the time complexity of Check If a Year Is a Leap Year in Python?

The optimal solution runs in O(1) time and O(1) auxiliary space. Three modulo checks, regardless of the year.

What follow-up questions do interviewers ask about Check If a Year Is a Leap Year?

Count the leap years between two years in O(1).