LeetCode Solutions Blog

Number of Days in a Month

Number: 1088

Difficulty: Easy

Paid? Yes

Companies: Amazon

Problem Description

Given a year and a month, determine the number of days in that month. Note that February can have 28 or 29 days, depending on whether the given year is a leap year.

Key Insights

Use the standard list of months with 31 and 30 days.
Special handling is required for February:
- A leap year is divisible by 4, but not by 100 unless it is also divisible by 400.
The solution uses constant time and space complexities, as it only involves conditional checks.

Space and Time Complexity

Time Complexity: O(1)
Space Complexity: O(1)

Solution

To solve the problem, we first check if the month is February. If it is, we determine if the year is a leap year using the given conditions. If it is a leap year, February has 29 days; otherwise, it has 28 days. For other months, we can use a mapping or conditions to output 31 days for months: January, March, May, July, August, October, and December; and 30 days for the remaining months: April, June, September, and November. This approach uses simple conditional checks without the need for additional data structures.

Code Solutions

# Function to determine the number of days in a given month of a given year.
def number_of_days(year: int, month: int) -> int:
    # Check if the month is February
    if month == 2:
        # Check for leap year: divisible by 4 and (not divisible by 100 or divisible by 400)
        if (year % 4 == 0 and (year % 100 != 0 or year % 400 == 0)):
            return 29  # Leap year February has 29 days
        else:
            return 28  # Non-leap year February has 28 days
    # Months with 31 days: January, March, May, July, August, October, December
    if month in {1, 3, 5, 7, 8, 10, 12}:
        return 31
    # Remaining months have 30 days: April, June, September, November
    return 30

# Example test cases
print(number_of_days(1992, 7))  # Expected output: 31
print(number_of_days(2000, 2))  # Expected output: 29
print(number_of_days(1900, 2))  # Expected output: 28