Problem Description

The task is to write a program that calculates the number of days between two given dates. The dates are provided in the string format 'YYYY-MM-DD', indicating the year, month, and day. The main challenge is to correctly handle the variations in the number of days in each month, leap years, and the possibility that the dates span multiple years. The solution requires attention to the rules that govern the Gregorian calendar, which include correct number of days in each month, leap year calculations, and the accumulation of days across multiple months and years.

Intuition

The intuition behind the provided solution lies in breaking down the task into smaller parts that are easier to manage:

1. Understand leap years: Leap years need special attention because they have an extra day in February (29 days instead of 28). The solution includes a function isLeapYear to check if a given year is a leap year. The rule for determining a leap year is that it should be divisible by 4 but not by 100 unless it is also divisible by 400.

2. Count days per month: The solution defines daysInMonth to handle the variation in the number of days per month, adjusting for leap years. An array holds the number of days in each month, with February being conditionally set to 28 or 29 days based on whether the year is a leap year.

3. Calculate days since a reference date: To calculate the total number of days of a date, calcDays processes each date independently, starting from a reference year (1971 in this implementation). It accumulates total days year by year and month by month up to the given date. Days in full years are counted by adding 365 for each year plus an additional day for each leap year. Days in months are counted by accumulating the days for each month leading up to the given month in the specified year.

4. Compute the absolute difference: Finally, the difference in days between the two dates is found by calculating the total days for each date since the reference date and subtracting the smaller one from the larger one to ensure a positive result. The abs function is used to return the absolute value of this difference.

This approach systematically adds up all the days from the reference date to each of the specified dates, and the difference between these sums represents the number of days between the dates.

Learn more about Math patterns.

Solution Approach

The implementation follows a straightforward algorithmic approach without using complex data structures or patterns. It focuses on accurately counting the days by considering each component of the date (year, month, day) one by one.

Here’s a step-by-step breakdown of the algorithm:

1. Leap Year Check: The function isLeapYear(year: int) -> bool determines whether a given year is a leap year. This is used to add an extra day for leap years when counting days.

   return year % 4 == 0 and (year % 100 != 0 or year % 400 == 0)

2. Days in Month Calculation: daysInMonth(year: int, month: int) -> int calculates the number of days in a given month for a specific year. It uses an array where each index corresponds to a month, with February’s days conditionally set to 29 if it is a leap year:

   days = [31, 28 + int(isLeapYear(year)), 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]
   return days[month - 1]

3. Total Days Calculator: calcDays(date: str) -> int converts a YYYY-MM-DD format date into the total number of days elapsed since a reference date, which in this code is January 1, 1971. It does this by first summing up all the days for the full years that have passed and then adding the days for the full months that have passed in the current year:

   year, month, day = map(int, date.split("-"))
   days = 0
   for y in range(1971, year):
       days += 365 + int(isLeapYear(y))
   for m in range(1, month):
       days += daysInMonth(year, m)
   days += day
   return days

4. Calculation of the Difference Between Dates: The days between the two dates are calculated by finding the absolute difference between the total days from the reference date to each input date:

   return abs(calcDays(date1) - calcDays(date2))

As you can see, the algorithm uses a combination of iteration and basic arithmetic operations to accurately calculate the total number of days between two dates. The problem can also be solved using Python's built-in datetime module, but this approach shows how it can be done manually, which could be beneficial in understanding how date calculations work or in environments where built-in date operations are not available.

Example Walkthrough

Let's consider two dates for this example: 2023-01-12 and 2023-03-04. Our task is to find the number of days between these dates using the steps outlined in the solution approach.

1. Leap Year Check: We first check if the years for both dates are leap years. Since we're only considering dates within 2023 for this example, this step is simple:

   isLeapYear(2023) # returns False, because 2023 is not divisible by 4

   Since 2023 is not a leap year, February will have 28 days.

2. Days in Month Calculation: We then calculate the number of days in each month up to the month before the given date for the year 2023:

   daysInMonth(2023, 1) # returns 31 days for January
   daysInMonth(2023, 2) # returns 28 days for February, as it's not a leap year

   January has 31 days, and February, being in a non-leap year, has 28 days.

3. Total Days Calculator: Now, let's convert each date to the total number of days elapsed since the reference date of January 1, 1971:

   calcDays("2023-01-12") # we sum the days from 1971 to 2022, plus days in Jan 2023, and then add 12
   calcDays("2023-03-04") # we sum the days from 1971 to 2022, plus days in Jan and Feb 2023, and then add 4

   Without going through all the calculations step by step, let's assume the function calcDays returns the total days correctly according to the algorithm.

4. Calculation of the Difference Between Dates: Finally, to find the difference in days between the two dates, we subtract the total days for 2023-01-12 from the total days for 2023-03-04 and take the absolute value:

   abs(calcDays("2023-03-04") - calcDays("2023-01-12"))

   Suppose calcDays("2023-01-12") returns 19044 and calcDays("2023-03-04") returns 19095, then our calculation will be:

   abs(19095 - 19044) # which gives us 51 days

   Based on our steps, there are 51 days between 2023-01-12 and 2023-03-04.

Through this small example, we illustrated how each step of the solution contributes to the final count of the days between two dates without relying on any external date-related libraries.

Solution Implementation

Time and Space Complexity

The given Python code calculates the number of days between two dates. We'll analyze the time complexity and space complexity separately.

Time Complexity:

1. The isLeapYear function is O(1), as it consists of only arithmetic operations.
2. The daysInMonth function is O(1) since it simply returns a value from a predefined list after calculating the leap year condition.
3. The calcDays function is the main function that determines time complexity:
   • It involves two for-loops:
     • The first loop runs from 1971 to the given year (excluding). If the given year Y, this first loop runs in O(Y−1971).
     • The second loop runs from 1 to the given month (excluding). Since the number of months is constant, this second loop runs in O(1).
   • The rest of the operations in calcDays are constant time calculations.
   • Assuming Y1 and Y2 are the respective years in date1 and date2, the total time complexity for invoking calcDays for both dates is O(Y1-1971 + Y2-1971).
4. The daysBetweenDates function calls calcDays twice and calculates the absolute difference between the returned values, which is an O(1) operation.
5. Combining all the above, the overall time complexity is O(Y1-1971 + Y2-1971).

Space Complexity:

1. The space used by the algorithm is constant since only a fixed amount of extra space is needed for the variables and arrays regardless of the input:
   • The days array in daysInMonth, which has constant size of 12.
   • The integer variables used throughout the functions for calculations.
2. There is no variable space allocation that depends on the size of the input, so the space complexity is O(1).

Learn more about how to find time and space complexity quickly using problem constraints.