Number of Days in a Month - Practice Coding Problems

Number of Days in a Month - Problem

Math Easy

Welcome to your role as a calendar system developer! You need to build a function that accurately determines the number of days in any given month of any year.

Given two integers year and month, return the exact number of days in that month. This might seem simple, but there's a catch - you need to handle leap years correctly!

Remember: February has 28 days in regular years, but 29 days in leap years. A leap year occurs every 4 years, except for years divisible by 100 (unless they're also divisible by 400).

Examples: January 2023 has 31 days, February 2020 (leap year) has 29 days, February 2021 (not leap year) has 28 days.

Input & Output

example_1.py — Regular month

$ Input: year = 2023, month = 1

› Output: 31

💡 Note: January always has 31 days regardless of the year

example_2.py — Leap year February

$ Input: year = 2020, month = 2

› Output: 29

💡 Note: 2020 is a leap year (divisible by 4), so February has 29 days

example_3.py — Non-leap year February

$ Input: year = 2021, month = 2

› Output: 28

💡 Note: 2021 is not a leap year (not divisible by 4), so February has 28 days

Constraints

1 ≤ month ≤ 12
1583 ≤ year ≤ 10⁴
Follow Gregorian calendar rules

Visualization

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Understanding the Visualization

Month Classification

Identify if the month is February (variable) or others (fixed)

Leap Year Detection

For February, apply the leap year formula

Direct Calculation

Return the appropriate number of days instantly

Key Takeaway

🎯 Key Insight: The optimal solution recognizes that only February has variable days - all other months are constants. This reduces the problem to a simple lookup with one special case handling.

Asked in

G Google 15 ⊞ Microsoft 12 a Amazon 8 Apple 6

The optimal solution uses a direct lookup approach with O(1) time complexity. Store month lengths in an array and handle February as a special case by checking leap year rules. A year is a leap year if it's divisible by 4, except for century years which must be divisible by 400.

Common Approaches

Approach	Time	Space	Notes
✓ Direct Lookup (Optimal)	O(1)	O(1)	Use lookup table for month lengths and calculate leap years directly
Brute Force (Day-by-Day Simulation)	O(1)	O(1)	Simulate each day of the month until we reach an invalid date

Direct Lookup (Optimal) — Algorithm Steps

Create an array with days for each month (assume February has 28)
If month is February, check if year is a leap year
Apply leap year rule: divisible by 4, but not by 100, unless also by 400
Return the appropriate number of days

Visualization

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Step-by-Step Walkthrough

Check Month

Determine if the month is February (special case)

Leap Year Check

If February, apply leap year rules

Direct Lookup

For other months, use lookup table

Return Result

Return the calculated number of days

Code -

solution.c — C

#include <stdio.h>

int numberOfDaysInMonth(int year, int month) {
    int days[] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
    
    if (month == 2) {
        if ((year % 4 == 0 && year % 100 != 0) || (year % 400 == 0)) {
            return 29;
        } else {
            return 28;
        }
    }
    
    return days[month - 1];
}

int main() {
    int year, month;
    scanf("%d %d", &year, &month);
    printf("%d\n", numberOfDaysInMonth(year, month));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(1)

Direct array lookup and simple arithmetic operations

✓ Linear Growth

Space Complexity

O(1)

Fixed size array and a few variables

✓ Linear Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Direct Lookup (Optimal) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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