What is the optimal time complexity of solving Leap year or not?

The optimized solution achieves a time complexity of O(n) by using sorting, mathematical shortcuts, or hash table lookups.

What is the auxiliary space complexity for Leap year or not?

The space complexity is O(1) since we only use a minimal/constant amount of extra memory for tracking variables (or auxiliary space if dynamic structures are allocated).

What are the typical edge cases we must handle in Leap year or not?

Key edge cases include empty collections, negative or large bounds causing integer overflow, arrays with only one element, or inputs where target criteria are not met.

How can we optimize the brute force approach for Leap year or not?

Brute force methods often inspect all possible combinations. We optimize this by sorting the input list to enable binary search, or tracking processed items in a set to avoid redundant nested loop lookups.

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Python BasicsEasy

Leap year or not

Learn how to solve the 'Leap year or not' problem. This detailed resource details brute force and optimized approaches.

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Problem Statement

Easy

Write a function is_leap_year(year) that takes an integer year and returns True if the year is a leap year, or False otherwise. A year is a leap year if it is divisible by 4 but not by 100, except when it is also divisible by 400.

Constraints

•1 <= year <= 10^5

Examples

Example 1

Input

year = 2024

Output

True

Explanation

2024 is divisible by 4 and not by 100, so it is a leap year.

Example 2

Input

year = 1900

Output

False

Explanation

1900 is divisible by 100 but not by 400, so it is not a leap year.

Example 3

Input

year = 2000

Output

True

Explanation

2000 is divisible by 400, so it is a leap year.

Need a Hint?

Use simple arithmetic operators (like modulo `%`, division `//`), conditional checks, or loops to inspect number properties.

Edge Cases to Watch

Empty list or null input variables
Single item lists/arrays
Extremely large input bounds causing integer or stack overflow

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