Solution Explanation: Day of the Year

This problem involves calculating the day of the year given a date in YYYY-MM-DD format. The solution leverages the Gregorian calendar rules for leap years and the number of days in each month.

Approach:

1. Parse the Date: Extract the year, month, and day from the input string.

2. Determine Leap Year: Check if the year is a leap year using the Gregorian calendar rules:

   • A year is a leap year if it's divisible by 400, or divisible by 4 but not by 100.

3. Calculate Days: Determine the number of days in each month. February requires special handling based on whether it's a leap year.

4. Cumulative Sum: Calculate the day of the year by summing the days of the preceding months and adding the current day.

Time Complexity: O(1) - The operations performed are constant time regardless of the input date.

Space Complexity: O(1) - Constant extra space is used to store variables and possibly a small array for the days in each month.

Code Implementation (Python):

class Solution:
    def dayOfYear(self, date: str) -> int:
        y, m, d = map(int, date.split('-'))  # Efficiently parse the date string
 
        # Determine leap year
        is_leap = (y % 400 == 0) or (y % 4 == 0 and y % 100 != 0)
 
        # Days in each month (handle leap year for February)
        days_in_month = [0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]
        if is_leap:
            days_in_month[2] = 29
 
        # Calculate day of the year
        day_of_year = d
        for i in range(1, m):
            day_of_year += days_in_month[i]
 
        return day_of_year
 

Code Implementation (Java):

class Solution {
    public int dayOfYear(String date) {
        String[] parts = date.split("-");
        int year = Integer.parseInt(parts[0]);
        int month = Integer.parseInt(parts[1]);
        int day = Integer.parseInt(parts[2]);
 
        boolean isLeap = (year % 400 == 0) || (year % 4 == 0 && year % 100 != 0);
        int[] daysInMonth = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
        if (isLeap) {
            daysInMonth[2] = 29;
        }
 
        int dayOfYear = day;
        for (int i = 1; i < month; i++) {
            dayOfYear += daysInMonth[i];
        }
        return dayOfYear;
    }
}

Other Languages: The approach remains the same across different languages; only the syntax and standard library functions for string manipulation and integer parsing might change. The provided solution includes implementations in C++, Go, TypeScript, and JavaScript, demonstrating this consistency. The core logic of leap year determination and cumulative day calculation remains unchanged.