0874EasyMathString

Day Number Within the Year

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Trains the technique from

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

A log stamp is given as date, a string in the form "YYYY-MM-DD".

Return which day of its own year that date is, counting the first of January as day 1.

A year has 366 days when it is a leap year and 365 otherwise. A year is a leap year when it divides by 4, except that a year dividing by 100 is not a leap year unless it also divides by 400.

Examples

Example 1

Input: date = "2020-12-31"
Output: 366

The year 2020 was a leap year, so it held 366 days and its final day is day 366.

Example 2

Input: date = "2020-03-01"
Output: 61

The year 2020 divides by 4 and not by 100, so February held 29 days, putting the first of March on day 61.

Example 3

Input: date = "1900-03-01"
Output: 60

The year 1900 divides by 100 but not by 400, so it was not a leap year and February held 28 days.

Constraints

date.length == 10
date[4] and date[7] are both '-', and every other character is a digit
date names a real calendar date between the first of January 1900 and the last of December 2100

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def day_of_the_year(date: str) -> int:

Java

public int dayOfTheYear(String date)

The intended approach

Time O(1): a fixed twelve-month walk.Space O(1).

Read the year, month and day out of their fixed positions in the stamp. Hold the ordinary lengths of the twelve months and add up the ones before the given month, then add the day. The only correction is February, so add one more when the month is past February in a leap year, using the divide-by-4 rule with its century exceptions.

Pythonentry point: day_of_the_year

def day_of_the_year(date):
    year = int(date[0:4])
    month = int(date[5:7])
    day = int(date[8:10])
    lengths = [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]
    total = day
    for m in range(month - 1):
        total += lengths[m]
    leap = year % 4 == 0 and (year % 100 != 0 or year % 400 == 0)
    if leap and month > 2:
        total += 1
    return total

The obvious version

Start at the first of January of that year and step forward one day at a time, rolling the month over when it runs out of days, counting the steps until the stamp is reached.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_day_of_the_year(date):
    year = int(date[0:4])
    month = int(date[5:7])
    day = int(date[8:10])
    leap = year % 4 == 0 and (year % 100 != 0 or year % 400 == 0)
    lengths = [31, 29 if leap else 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]
    count = 0
    m = 1
    d = 1
    while True:
        count += 1
        if m == month and d == day:
            return count
        d += 1
        if d > lengths[m - 1]:
            d = 1
            m += 1

How this problem was checked

Reference solution executed against 12 test cases — 12 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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