CF 102443I - Dates

Each input line describes one date, but the order of its components depends on the separator. A dot means the European-style order day.month.year, while a slash means the American-style order month/day/year. The task is not to decide whether the date is a real calendar date.

codeforcescompetitive-programming

CF 102443I - Dates

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Solve time: 10m 25s
Verified: yes

Solution

Problem Understanding

Each input line describes one date, but the order of its components depends on the separator. A dot means the European-style order day.month.year, while a slash means the American-style order month/day/year. The task is not to decide whether the date is a real calendar date. Even something such as 31.02.2001 must still be converted and printed.

For every date, we must produce both representations. The first always has the order day, month, year and uses dots. The second always has month, day, year and uses slashes. The day and month must occupy exactly two digits, while the year must occupy exactly four digits. Leading zeroes in the input can be removed during parsing and then restored to the required width in the output.

There are at most 20,000 input dates. Each date contains only three small numeric fields, so there is no reason to use any data structure more complicated than a few strings or integers per line. An algorithm that processes each date once is easily fast enough. Even a constant amount of extra work per date is negligible at this scale. The 512 MB memory limit is also far beyond what is needed, since the dates can be processed independently without storing the entire input.

The first subtle case is a one-digit year. For example,

1
1.2.1

must become

01.02.0001 02/01/0001

A careless implementation using the input strings directly might print 01.02.1, because the input permits a short year while the output always requires four digits.

A second edge case is an already padded date:

1
01.02.0001

The output is still

01.02.0001 02/01/0001

The conversion must not accidentally treat the leading zeroes as part of the numeric value or produce inconsistent widths.

A third case is an invalid calendar date:

1
29.02.2001

The required output is

29.02.2001 02/29/2001

The year 2001 is not a leap year, but validity is irrelevant to the task. Adding calendar validation would solve a problem that was never asked.

A fourth case exercises the maximum ordinary values:

1
31/12/9999

The result is

31.12.9999 12/31/9999

Here the separator must determine the original ordering correctly, and the year must remain four digits without any special handling.

Approaches

A deliberately brute-force interpretation would treat the three parsed components as an unordered collection and try every possible ordering until one matches the required format. There are only 3! = 6 permutations, so for each of the 20,000 dates this would perform at most 120,000 permutation attempts, plus formatting work. It is still fast enough for these constraints, but it solves a harder problem than necessary and makes the code more complicated. More importantly, the separator already tells us the ordering directly, so there is nothing to search for.

The brute-force approach works because there are only six possible arrangements of three fields, but fails as an engineering choice because all six possibilities are unnecessary. The observation that the input separator uniquely identifies the format reduces the problem to one deterministic split and one swap.

For a dot-separated date, the fields are already in day, month, year order. For a slash-separated date, the first two fields are month and day, so they only need to be exchanged. Once the three fields are in day, month, year order, formatting is mechanical: pad day and month to two characters and year to four.

There is no need to validate month lengths, leap years, or even whether the resulting combination is a real date. The input guarantees that each component lies within its stated numeric range, and the output is simply a normalized representation of those components.

Approach Time Complexity Space Complexity Verdict
Brute Force O(6n) = O(n) O(1) Accepted, but unnecessary work
Optimal O(n) O(1) auxiliary space Accepted

Algorithm Walkthrough

  1. Read one date string and inspect its separator. A dot means the fields are ordered as day, month, year, while a slash means they are ordered as month, day, year. The separator gives us the format without any ambiguity.
  2. Split the string into its three components. We keep the components as strings because the only transformation needed is padding, and strings avoid unnecessary integer conversion.
  3. If the separator is a dot, assign the three fields directly to day, month, and year. If the separator is a slash, assign the first field to month, the second to day, and the third to year.
  4. Pad day and month to exactly two characters and year to exactly four characters. Padding rather than manually checking the number of digits handles both already-padded and unpadded input uniformly.
  5. Construct the European representation as DD.MM.YYYY.
  6. Construct the American representation as MM/DD/YYYY.
  7. Print the two representations separated by one space. Each input line is independent, so the same procedure can be repeated immediately for the next date.

Why it works

After step 3, the three variables always represent the same logical date in the canonical order day, month, year, regardless of how the input was written. The separator determines which assignment is correct, so no other interpretation is possible. Padding changes only the textual width of each component, not its value. Consequently, the first constructed string always places the canonical components in day-month-year order with the required widths, while the second places the same components in month-day-year order. Since no calendar validation is required, every permitted input is handled correctly.

Python Solution

import sys
input = sys.stdin.readline

def solve():
    n = int(input())

    for _ in range(n):
        s = input().strip()

        if '.' in s:
            day, month, year = s.split('.')
        else:
            month, day, year = s.split('/')

        day = day.zfill(2)
        month = month.zfill(2)
        year = year.zfill(4)

        european = f"{day}.{month}.{year}"
        american = f"{month}/{day}/{year}"

        print(european, american)

if __name__ == "__main__":
    solve()

The separator check implements the first decision in the algorithm. Since the input has exactly one of the two permitted separators, checking for '.' is sufficient. The alternative case must be slash-separated.

The assignment order in the two branches is the key part of the solution. For 11.12.2000, splitting produces ["11", "12", "2000"], which already means day, month, year. For 1/29/3000, splitting produces ["1", "29", "3000"], which means month, day, year, so the first two variables are deliberately assigned in the opposite order.

zfill handles every permitted width without arithmetic. zfill(2) leaves "11" unchanged and converts "1" to "01". Similarly, zfill(4) converts "1" to "0001" and "2100" remains "2100". This also avoids an easy mistake where only the day and month are padded while the year is left at its input width.

No integer overflow is possible because the implementation never performs arithmetic on the date components. It also does not check whether a date is valid, which is correct because invalid dates must still be formatted.

Worked Examples

Sample 1

The first input date is dot-separated, so its components already have day-month-year ordering. The second date has the same ordering but contains one-digit day, month, and year fields.

Input Separator Day Month Year European American
11.12.2000 . 11 12 2000 11.12.2000 12/11/2000
1.2.1 . 1 2 1 01.02.0001 02/01/0001

The second row demonstrates why output padding must be independent of input padding. The input year has one digit, but the normalized representation has four digits.

Sample 2

The first date is again dot-separated. The second is slash-separated, so the first two components must be exchanged when assigning day and month.

Input Separator Day Month Year European American
20.10.2100 . 20 10 2100 20.10.2100 10/20/2100
1/29/3000 / 29 1 3000 29.01.3000 01/29/3000

The second row confirms the central invariant: after parsing, day, month, and year have the same meaning regardless of the original notation.

Complexity Analysis

Measure Complexity Explanation
Time O(n) Every date is split and formatted once, with only three constant-size fields.
Space O(1) auxiliary Only the current date and its three components are kept.

With at most 20,000 dates, the algorithm performs a small constant amount of string processing per line. It is comfortably within the 1 second time limit and uses negligible memory compared with the 512 MB limit.

Test Cases

import sys
import io

def solve():
    input = sys.stdin.readline
    n = int(input())

    out = []

    for _ in range(n):
        s = input().strip()

        if '.' in s:
            day, month, year = s.split('.')
        else:
            month, day, year = s.split('/')

        day = day.zfill(2)
        month = month.zfill(2)
        year = year.zfill(4)

        out.append(f"{day}.{month}.{year} {month}/{day}/{year}")

    sys.stdout.write("\n".join(out))

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

assert run(
    "2\n"
    "11.12.2000\n"
    "1.2.1\n"
) == (
    "11.12.2000 12/11/2000\n"
    "01.02.0001 02/01/0001"
), "sample 1"

assert run(
    "2\n"
    "20.10.2100\n"
    "1/29/3000\n"
) == (
    "20.10.2100 10/20/2100\n"
    "29.01.3000 01/29/3000"
), "sample 2"

assert run(
    "1\n"
    "1.1.1\n"
) == (
    "01.01.0001 01/01/0001"
), "minimum values"

assert run(
    "4\n"
    "31.12.9999\n"
    "31/12/9999\n"
    "01.01.0001\n"
    "01/01/0001\n"
) == (
    "31.12.9999 12/31/9999\n"
    "31.12.9999 12/31/9999\n"
    "01.01.0001 01/01/0001\n"
    "01.01.0001 01/01/0001"
), "boundaries and padded values"

assert run(
    "3\n"
    "29.02.2001\n"
    "31/02/2001\n"
    "30.04.2000\n"
) == (
    "29.02.2001 02/29/2001\n"
    "02.31.2001 02/31/2001\n"
    "30.04.2000 04/30/2000"
), "invalid calendar dates are still formatted"

assert run(
    "5\n"
    "7.7.7\n"
    "7/7/7\n"
    "07.07.07\n"
    "07/07/07\n"
    "0007.01.0007\n"
) == (
    "07.07.0007 07/07/0007\n"
    "07.07.0007 07/07/0007\n"
    "07.07.0007 07/07/0007\n"
    "07.07.0007 07/07/0007\n"
    "07.01.0007 01/07/0007"
), "padding and repeated values"

The minimum-value case checks that every field receives the required leading zeroes. The boundary case checks both ends of the permitted ranges and confirms that already padded fields remain unchanged. The invalid-date case catches an implementation that incorrectly tries to validate the calendar before formatting. The final case exercises short and already padded fields in both formats, including a year with leading zeroes.

Test input Expected output What it validates
1\n1.1.1 01.01.0001 01/01/0001 Minimum values and all required padding
4\n31.12.9999\n31/12/9999\n01.01.0001\n01/01/0001 Corresponding normalized dates Boundary values and both separators
3\n29.02.2001\n31/02/2001\n30.04.2000 Components reformatted without validation Invalid calendar dates
5\n7.7.7\n7/7/7\n07.07.07\n07/07/07\n0007.01.0007 All components normalized to fixed widths Leading zeroes and mixed input widths

Edge Cases

One-digit year

For the input

1
1.2.1

the separator is a dot, so the components are day = "1", month = "2", and year = "1". Padding produces 01, 02, and 0001. The two outputs are consequently 01.02.0001 and 02/01/0001. The algorithm never assumes that the input year already has four characters.

Already padded values

For

1
01.02.0001

the split produces fields that already have the desired widths. Calling zfill does not add anything because zfill leaves strings at or above the requested width unchanged. The output is

01.02.0001 02/01/0001

This makes the normalization idempotent for correctly formatted input.

Invalid calendar date

For

1
29.02.2001

the algorithm does not inspect the number of days in February. It simply parses 29, 02, and 2001, pads them, and constructs both representations. The result is

29.02.2001 02/29/2001

This is exactly what the problem requires, even though the date is not a valid Gregorian calendar date.

Slash-separated input

For

1
1/29/3000

the separator is /, so the parser assigns month = "1", day = "29", and year = "3000". After padding, the values are 01, 29, and 3000. The European form becomes 29.01.3000, while the American form becomes 01/29/3000. This case catches the most likely parsing error, which is treating slash-separated input as if it were already in day-month-year order.