Day of the Year

Day of the Year - Problem

Math String Easy

Given a string date representing a Gregorian calendar date formatted as YYYY-MM-DD, return the day number of the year.

The day number represents which day of the year it is. For example, January 1st is day 1, January 2nd is day 2, and so on.

Note: You need to consider leap years when February has 29 days instead of 28.

Input & Output

Example 1 — Early February

$ Input: date = "2019-02-10"

› Output: 41

💡 Note: January has 31 days, so February 10th is the 31 + 10 = 41st day of the year

Example 2 — Leap Year March

$ Input: date = "2020-03-01"

› Output: 61

💡 Note: 2020 is a leap year. January (31) + February (29) + March 1st = 31 + 29 + 1 = 61

Example 3 — Year Start

$ Input: date = "2021-01-01"

› Output: 1

💡 Note: January 1st is always the 1st day of any year

Constraints

date.length == 10
date[4] == date[7] == '-'
All other characters in date are digits
date represents a calendar date between Jan 1st, 1900 and Dec 31st, 2019

Visualization

Tap to expand

Asked in

M Microsoft 15 Apple 8

The key insight is to use cumulative day counts for each month and handle leap years properly. Best approach is prefix sum array with O(1) lookup. Time: O(1), Space: O(1)

Common Approaches

✓ Day-by-Day Counting

⏱️ Time: O(m) Space: O(1)

Iterate through each month and each day, incrementing a counter until we reach the target date. Handle leap years by checking if February should have 29 days.

Prefix Sum Array

⏱️ Time: O(1) Space: O(1)

Create an array storing cumulative days up to each month, then add the day number. This eliminates the need to loop through months every time.

Day-by-Day Counting — Algorithm Steps

Parse year, month, and day from input string
Count days month by month from January
Add remaining days in target month

Visualization

Tap to expand

Step-by-Step Walkthrough

Parse Date

Extract year=2019, month=2, day=10

Count January

Add 31 days for complete January

Add February Days

Add 10 days from February, total = 41

Code -

solution.c — C

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

int solution(char* date) {
    int year, month, day;
    sscanf(date, "%d-%d-%d", &year, &month, &day);
    
    // Days in each month (non-leap year)
    int daysInMonth[] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
    
    // Check if it's a leap year
    int isLeap = (year % 4 == 0 && year % 100 != 0) || (year % 400 == 0);
    if (isLeap) {
        daysInMonth[1] = 29;
    }
    
    int totalDays = 0;
    
    // Count days in complete months
    for (int i = 0; i < month - 1; i++) {
        totalDays += daysInMonth[i];
    }
    
    // Add days in current month
    totalDays += day;
    
    return totalDays;
}

int main() {
    char date[12];
    fgets(date, sizeof(date), stdin);
    date[strcspn(date, "\n")] = 0;
    printf("%d\n", solution(date));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(m)

Where m is the month number (1-12), as we iterate through each month

✓ Linear Growth

Space Complexity

O(1)

Only using a few variables to track counts

✓ Linear Space

24.0K Views

Medium Frequency

~15 min Avg. Time

890 Likes

Ln 1, Col 1

Smart Actions

💡 Explanation

AI Ready

💡 Suggestion Tab to accept Esc to dismiss

// Output will appear here after running code

Code Editor Closed

Click the red button to reopen

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Day-by-Day Counting — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Select Compiler