Number of Days in a Month - Practice Coding Problems

Number of Days in a Month - Problem

Math Easy

Given a year and a month, return the number of days in that month.

You need to handle leap years correctly. A leap year occurs:

Every 4 years, BUT
NOT every 100 years, BUT
Every 400 years

For example, 2000 and 2004 are leap years, but 1900 is not.

Remember that February has 28 days in regular years and 29 days in leap years.

Input & Output

Example 1 — Leap Year February

$ Input: year = 2020, month = 2

› Output: 29

💡 Note: 2020 is a leap year (divisible by 4, not by 100), so February has 29 days

Example 2 — Regular Month

$ Input: year = 2019, month = 8

› Output: 31

💡 Note: August always has 31 days regardless of leap year

Example 3 — Non-Leap February

$ Input: year = 1900, month = 2

› Output: 28

💡 Note: 1900 is not a leap year (divisible by 100 but not by 400), so February has 28 days

Constraints

1583 ≤ year ≤ 2100
1 ≤ month ≤ 12

Visualization

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Understanding the Visualization

Input

Year and month values

Logic

Check leap year for February, use fixed values for others

Output

Number of days in that month

Key Takeaway

🎯 Key Insight: February is the only month affected by leap years - use modulo arithmetic to check leap year conditions efficiently

Asked in

G Google 15 M Microsoft 12 a Amazon 8

The key insight is that only February varies based on leap years - all other months have fixed day counts. The optimal approach uses either an array lookup with leap year checking or mathematical patterns. Time: O(1), Space: O(1)

Common Approaches

Approach	Time	Space	Notes
✓ Mathematical Formula	O(1)	O(1)	Use mathematical patterns to determine days without arrays
Array Lookup	O(1)	O(1)	Use predefined array with days per month and leap year helper
If-Else Chain	O(1)	O(1)	Check each month with individual if-else conditions

Mathematical Formula — Algorithm Steps

Step 1: Check if month is February, apply leap year formula
Step 2: Use mathematical pattern for 31-day months
Step 3: Default remaining months to 30 days

Visualization

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Step-by-Step Walkthrough

Pattern Split

Divide months into two groups: ≤7 and >7

Apply Modulo

Use odd/even pattern with modulo 2

Handle February

Special case for leap year calculation

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int solution(int year, int month) {
    // February: check leap year
    if (month == 2) {
        return ((year % 400 == 0) || (year % 4 == 0 && year % 100 != 0)) ? 29 : 28;
    }
    
    // Months with 31 days: 1,3,5,7,8,10,12
    if (month <= 7) {
        return (month % 2 == 1) ? 31 : 30;
    } else {
        return (month % 2 == 0) ? 31 : 30;
    }
}

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

Time & Space Complexity

Time Complexity

⏱️

O(1)

Direct mathematical calculations with no loops

✓ Linear Growth

Space Complexity

O(1)

No extra data structures, only variables

✓ Linear Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Mathematical Formula — Algorithm Steps

Visualization

Code -

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