CF 102697054 - When is my meeting?

We receive two calendar dates. The first date represents today, and the second date represents the scheduled meeting. Each date is written as a month name, a day number, and a year, such as July 5, 2019. The task is to determine the number of days separating the two dates.

codeforcescompetitive-programming

CF 102697054 - When is my meeting?

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Solution

Problem Understanding

We receive two calendar dates. The first date represents today, and the second date represents the scheduled meeting. Each date is written as a month name, a day number, and a year, such as July 5, 2019.

The task is to determine the number of days separating the two dates. If both dates are identical, the meeting is today. If the meeting date is earlier, we need to report how many days late the meeting is. Otherwise, we report how many days remain until the meeting. The statement explicitly says that leap years do not need to be considered.

There are no large arrays, graphs, or repeated test cases here. The input contains exactly two dates, so even a linear scan through a small fixed number of months is effectively constant time. The one-second limit is therefore more than sufficient. The main challenge is not performance but converting the human-readable date into a representation where two dates can be compared and subtracted safely.

The first edge case is equality. For example,

January 10, 2020January 10, 2020

produces

It is today

A careless implementation that only checks whether the meeting date is greater than today and otherwise treats it as a past date would incorrectly report that the meeting is one or zero days late.

The second edge case is a meeting in the next month. For example,

January 31, 2020February 1, 2020

has a difference of one day. Counting only the difference between the day numbers would produce -30, which completely ignores the month boundary.

The third edge case is a meeting in a previous month. For example,

March 1, 2020February 28, 2020

is two days late under the problem's simplified calendar, because February has 28 days and leap years are ignored. A solution that only compares the day fields would see 28 > 1 and could mistakenly conclude that the meeting is in the future.

The fourth edge case is a year boundary. For example,

December 31, 2019January 1, 2020

has a difference of one day. Comparing months and days without incorporating the year can easily reverse the ordering.

Approaches

A direct brute-force approach would start at the first date and repeatedly advance one day until it reaches the meeting date, or repeatedly move backward when the meeting has already happened. This is correct because every transition represents exactly one calendar day. However, its running time depends on the number of days separating the dates. Since the problem does not publish a numerical bound on the years in the input, that approach has no useful fixed upper bound and can require roughly 365 * |year_difference| iterations.

There is a much cleaner representation. Since the lengths of all months are known and leap years can be ignored, we can assign every date an absolute day number. A date's absolute position is the number of complete years before it, plus the days contributed by all complete months before it, plus its day within the month.

For example, if the date is March 10, then all days in January and February are already before it. Once both input dates have been converted into this representation, the entire calendar calculation becomes ordinary integer subtraction.

The brute force works because advancing one day at a time accurately models the calendar, but it fails because it performs one operation per day. The observation that every date can instead be mapped to an absolute day index lets us replace potentially thousands or millions of individual transitions with two conversions and one subtraction.

The month lengths can be stored as [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]. Because leap years are explicitly excluded, February always contributes 28 days.

Approach Time Complexity Space Complexity Verdict
Brute Force O(D) O(1) Too slow when dates are far apart
Optimal O(1) O(1) Accepted

Here D is the number of days between the two dates. The optimal solution examines only the twelve month lengths, which is constant work.

Algorithm Walkthrough

  1. Store the number of days in each month in calendar order. February is fixed at 28 because leap years are not part of this problem.
  2. Parse each input line into its month name, day, and year. The comma after the day is part of the textual format, so it is easiest to remove it before converting the day and year to integers.
  3. Convert each month name into its numerical month index. This lets us use the month-length array directly.
  4. Convert each date into an absolute day number. Start with all complete years before the date, contributing year * 365 days. Then add the lengths of all complete months before the current month and finally add the current day.
  5. Subtract the absolute day number of the first date from that of the meeting. A positive result means the meeting is in the future, zero means it is today, and a negative result means it has already passed.
  6. Format the answer according to the sign of the difference. For a positive difference d, print that there are d days until the meeting. For zero, print It is today. For a negative difference, use -d as the number of days late.

Why this works is captured by the absolute-day representation. Every ordinary year contributes exactly 365 days, and every month contributes its fixed number of days. Consequently, two dates that are one calendar day apart receive absolute day numbers differing by exactly one. The subtraction is therefore exactly the number of days separating the dates. Its sign also gives the chronological ordering, so the final message cannot select the wrong case.

Python Solution

Pythonimport sysinput = sys.stdin.readline
MONTHS = {    "January": 1,    "February": 2,    "March": 3,    "April": 4,    "May": 5,    "June": 6,    "July": 7,    "August": 8,    "September": 9,    "October": 10,    "November": 11,    "December": 12,}
DAYS_IN_MONTH = [    31, 28, 31, 30, 31, 30,    31, 31, 30, 31, 30, 31]

def parse_date(s):    month_name, rest = s.split()    day, year = rest.split(",")
    return MONTHS[month_name], int(day), int(year)

def to_day_number(date):    month, day, year = date
    total = year * 365
    for i in range(month - 1):        total += DAYS_IN_MONTH[i]
    total += day    return total

def solve():    first = input().strip()    meeting = input().strip()
    first_date = parse_date(first)    meeting_date = parse_date(meeting)
    first_day = to_day_number(first_date)    meeting_day = to_day_number(meeting_date)
    diff = meeting_day - first_day
    if diff == 0:        print("It is today")    elif diff > 0:        print(f"There are {diff} days until your meeting")    else:        print(f"Sorry you are {-diff} days late")

if __name__ == "__main__":    solve()

parse_date handles the exact textual structure of a date. Splitting July 5, 2019 gives July and 5, 2019, after which splitting the second part at the comma produces the numeric day and year.

to_day_number deliberately uses year * 365, rather than a leap-year-aware formula. This is not an approximation introduced by the solution, it is exactly the calendar model specified by the problem.

The loop over previous months uses range(month - 1). For January this executes zero times, which is the correct boundary condition because there are no complete months before January. For March it processes January and February, exactly the two complete months preceding March.

Python integers do not overflow, so there is no special handling needed for large years. The subtraction is performed only after both dates have been converted to the same coordinate system, which avoids separate cases for crossing months or years.

Worked Examples

The Codeforces page currently displays one sample, although its expected output conflicts with the formal Output section. The sample input is:

July 5, 2019March 10, 2020

Using the formal specification, the calculation proceeds as follows. July 5, 2019 is converted into its absolute day number, and March 10, 2020 is converted similarly. The difference is 249 days under the stated no-leap-year calendar? Let's calculate carefully: July 5 to December 31 is 26 + 31 + 30 + 31 + 30 + 31 = 179 days, and January 1 to March 10 is 31 + 28 + 10 = 69, giving 248 days. Thus the difference is 248.

Date Year contribution Previous months Day Absolute day
July 5, 2019 2019 × 365 181 5 737131
March 10, 2020 2020 × 365 59 10 737379

The difference is 737379 - 737131 = 248, so the formal output is:

There are 248 days until your meeting

The displayed sample instead says:

The meeting will occur in 248 days.

That discrepancy is present on the current Codeforces page itself.

A second useful example crosses a year boundary:

December 31, 2019January 1, 2020
Date Year contribution Previous months Day Absolute day
December 31, 2019 2019 × 365 334 31 737284
January 1, 2020 2020 × 365 0 1 737285

The difference is one, so the algorithm prints:

There are 1 days until your meeting

The wording is intentionally kept exactly as specified by the formal Output section, including its use of days for the value 1.

Complexity Analysis

Measure Complexity Explanation
Time O(1) Each date scans at most 11 previous months
Space O(1) Only a fixed month table and a constant number of variables are used

The input itself contains only two dates, and each conversion performs at most twelve month operations. With the published one-second limit and 256 MB memory limit, this is comfortably within the available resources.

Test Cases

Because the published statement does not give numeric bounds on the years, the maximum-size test below uses a large year value to exercise integer handling. Python's arbitrary-precision integers make this safe.

The helper calls the same solve function used by the submitted program, while temporarily replacing standard input and output so each case can be tested independently.

Python# helper: run solution on input string, return output stringimport sysimport io
MONTHS = {    "January": 1,    "February": 2,    "March": 3,    "April": 4,    "May": 5,    "June": 6,    "July": 7,    "August": 8,    "September": 9,    "October": 10,    "November": 11,    "December": 12,}
DAYS_IN_MONTH = [    31, 28, 31, 30, 31, 30,    31, 31, 30, 31, 30, 31]

def parse_date(s):    month_name, rest = s.split()    day, year = rest.split(",")    return MONTHS[month_name], int(day), int(year)

def to_day_number(date):    month, day, year = date    total = year * 365    for i in range(month - 1):        total += DAYS_IN_MONTH[i]    return total + day

def solve():    first = input().strip()    meeting = input().strip()
    first_day = to_day_number(parse_date(first))    meeting_day = to_day_number(parse_date(meeting))
    diff = meeting_day - first_day
    if diff == 0:        print("It is today")    elif diff > 0:        print(f"There are {diff} days until your meeting")    else:        print(f"Sorry you are {-diff} days late")

def run(inp: str) -> str:    old_stdin = sys.stdin    old_stdout = sys.stdout
    try:        sys.stdin = io.StringIO(inp)        sys.stdout = io.StringIO()        solve()        return sys.stdout.getvalue()    finally:        sys.stdin = old_stdin        sys.stdout = old_stdout

# Provided sample, interpreted according to the formal Output section.assert run(    "July 5, 2019\n"    "March 10, 2020\n") == "There are 248 days until your meeting\n", "sample 1"
# Minimum-size style case: identical dates.assert run(    "January 1, 1\n"    "January 1, 1\n") == "It is today\n", "same date"
# Year boundary.assert run(    "December 31, 2019\n"    "January 1, 2020\n") == "There are 1 days until your meeting\n", "year boundary"
# Meeting in the past, crossing a month boundary.assert run(    "March 1, 2020\n"    "February 28, 2020\n") == "Sorry you are 2 days late\n", "past meeting"
# All fields on a month boundary.assert run(    "January 31, 2020\n"    "February 1, 2020\n") == "There are 1 days until your meeting\n", "month boundary"
# Large years, exercising integer handling.assert run(    "January 1, 1000000000\n"    "December 31, 1000000000\n") == "There are 364 days until your meeting\n", "large year"
Test input Expected output What it validates
January 1, 1 / January 1, 1 It is today Equality and minimum-style input
December 31, 2019 / January 1, 2020 There are 1 days until your meeting Year boundary
March 1, 2020 / February 28, 2020 Sorry you are 2 days late Past date and month reversal
January 31, 2020 / February 1, 2020 There are 1 days until your meeting Month boundary and off-by-one handling
January 1, 1000000000 / December 31, 1000000000 There are 364 days until your meeting Large year arithmetic

Edge Cases

For equal dates, consider:

January 1, 1January 1, 1

Both dates receive exactly the same absolute day number. Their difference is zero, so the algorithm selects the equality branch and prints It is today. No special calendar calculation is needed for this case.

For a month boundary, consider:

January 31, 2020February 1, 2020

January contributes 31 days. The first date has absolute position year * 365 + 31, while the second has year * 365 + 31 + 1, so their difference is exactly one. This catches implementations that subtract the day fields directly.

For a past meeting, consider:

March 1, 2020February 28, 2020

March 1 has February's 28 days before it, while February 28 has only the first 27 February days before its own day position. The absolute positions differ by two, with the meeting's position smaller. The algorithm obtains diff = -2, negates it for the displayed magnitude, and prints Sorry you are 2 days late.

For a year boundary, consider:

December 31, 2019January 1, 2020

The first date is the final day represented by 2019, and the second is the immediately following day. The year contribution changes by 365 while the month contribution resets, but the absolute representation absorbs both changes automatically. The resulting difference is one.

Finally, February must always be treated as 28 days. The problem explicitly removes leap-year handling, so February 28, 2020 is followed by March 1, 2020 with a one-day gap in this problem's calendar model. Adding leap-year logic would introduce behavior that the statement does not request.