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Count the Days Between Two Dates | Optiver Coding Question

Quick Overview

This question evaluates date arithmetic, calendar logic, string parsing, and correct leap-year handling as implementation competencies. It is commonly asked to assess algorithmic correctness and edge-case handling across month, year, and leap boundaries, falls under the Coding & Algorithms domain (date/calendar problems), and targets practical application rather than purely conceptual understanding.

Count the Days Between Two Dates

Company: Optiver

Role: Software Engineer

Category: Coding & Algorithms

Difficulty: medium

Interview Round: Take-home Project

# Count the Days Between Two Dates You are given two calendar dates as strings in `YYYY-MM-DD` format. Return the number of days between them as a non-negative integer. You must compute the result **without using any built-in date/time or calendar library** (e.g., no `datetime`, `java.time`, `Date`, or similar). The only calendar knowledge you may rely on is the standard Gregorian calendar rules given below. Assume a helper that returns the number of days in a given month is something you write yourself. ## Calendar rules - Months have the usual lengths: January 31, February 28 (29 in a leap year), March 31, April 30, May 31, June 30, July 31, August 31, September 30, October 31, November 30, December 31. - A year is a **leap year** if it is divisible by 4, **except** years divisible by 100, **unless** they are also divisible by 400. So 2000 and 2024 are leap years; 1900 and 2100 are not. ## Input - `date1`: a string in `YYYY-MM-DD` format. - `date2`: a string in `YYYY-MM-DD` format. Both dates are guaranteed to be valid calendar dates between `1900-01-01` and `2100-12-31` inclusive. The dates may be given in either order (`date1` is not necessarily earlier than `date2`). ## Output Return a single integer: the absolute difference in days between the two dates. If the dates are equal, return `0`. ## Examples | date1 | date2 | Output | Explanation | |-------|-------|--------|-------------| | `2019-06-29` | `2019-06-30` | `1` | Consecutive days. | | `2019-12-31` | `2020-01-15` | `15` | Crosses a year boundary. | | `2020-02-28` | `2020-03-01` | `2` | 2020 is a leap year, so February 29 lies between them. | | `2019-02-28` | `2019-03-01` | `1` | 2019 is not a leap year. | | `2021-07-04` | `2021-07-04` | `0` | Same date. | ## Constraints - `1900-01-01 <= date1, date2 <= 2100-12-31` - Both inputs are valid dates in `YYYY-MM-DD` format. - No date/time library calls are allowed; leap-year handling must be implemented from the rules above.

Quick Answer: This question evaluates date arithmetic, calendar logic, string parsing, and correct leap-year handling as implementation competencies. It is commonly asked to assess algorithmic correctness and edge-case handling across month, year, and leap boundaries, falls under the Coding & Algorithms domain (date/calendar problems), and targets practical application rather than purely conceptual understanding.

You are given two calendar dates as strings in `YYYY-MM-DD` format. Return the number of days between them as a non-negative integer. You must compute the result **without using any built-in date/time or calendar library** (no `datetime`, `java.time`, `Date`, or similar). The only calendar knowledge you may rely on is the standard Gregorian calendar rules below; write your own helper for the number of days in a month. ## Calendar rules - Month lengths: Jan 31, Feb 28 (29 in a leap year), Mar 31, Apr 30, May 31, Jun 30, Jul 31, Aug 31, Sep 30, Oct 31, Nov 30, Dec 31. - A year is a **leap year** if it is divisible by 4, **except** years divisible by 100, **unless** they are also divisible by 400. So 2000 and 2024 are leap years; 1900 and 2100 are not. ## Input - `date1`: a string in `YYYY-MM-DD` format. - `date2`: a string in `YYYY-MM-DD` format. Both dates are valid calendar dates between `1900-01-01` and `2100-12-31` inclusive, and may be given in either order (`date1` is not necessarily earlier than `date2`). ## Output Return a single integer: the absolute difference in days between the two dates. If the dates are equal, return `0`. ## Examples | date1 | date2 | Output | |-------|-------|--------| | `2019-06-29` | `2019-06-30` | `1` | | `2019-12-31` | `2020-01-15` | `15` | | `2020-02-28` | `2020-03-01` | `2` | | `2019-02-28` | `2019-03-01` | `1` | | `2021-07-04` | `2021-07-04` | `0` |

Constraints

1900-01-01 <= date1, date2 <= 2100-12-31
Both inputs are valid dates in YYYY-MM-DD format.
date1 is not necessarily earlier than date2.
No date/time library calls are allowed; leap-year handling must be implemented from the Gregorian rules.

Examples

Input: ('2019-06-29', '2019-06-30')

Expected Output: 1

Explanation: Consecutive days.

Input: ('2019-12-31', '2020-01-15')

Expected Output: 15

Explanation: Crosses a year boundary.

Hints

Convert each date to a single integer 'day count' from a fixed anchor (e.g. days since 1900-01-01), then the answer is the absolute difference of the two counts.
To count days into a year, sum days_in_month for all months before the given month, then add the day-of-month. Sum whole-year lengths (365 or 366) for the years before the given year.
A year is a leap year if (year % 4 == 0 and year % 100 != 0) or year % 400 == 0. Test 1900 (not leap), 2000 (leap), 2024 (leap).
Since the dates may be in either order, take abs() of the difference so the result is always non-negative.