Python/project_euler/problem_191/sol1.py

"""
Prize Strings
Problem 191

A particular school offers cash rewards to children with good attendance and
punctuality. If they are absent for three consecutive days or late on more
than one occasion then they forfeit their prize.

During an n-day period a trinary string is formed for each child consisting
of L's (late), O's (on time), and A's (absent).

Although there are eighty-one trinary strings for a 4-day period that can be
formed, exactly forty-three strings would lead to a prize:

OOOO OOOA OOOL OOAO OOAA OOAL OOLO OOLA OAOO OAOA
OAOL OAAO OAAL OALO OALA OLOO OLOA OLAO OLAA AOOO
AOOA AOOL AOAO AOAA AOAL AOLO AOLA AAOO AAOA AAOL
AALO AALA ALOO ALOA ALAO ALAA LOOO LOOA LOAO LOAA
LAOO LAOA LAAO

How many "prize" strings exist over a 30-day period?

References:
    - The original Project Euler project page:
      https://projecteuler.net/problem=191
"""

cache: dict[tuple[int, int, int], int] = {}


def _calculate(days: int, absent: int, late: int) -> int:
    """
    A small helper function for the recursion, mainly to have
    a clean interface for the solution() function below.

    It should get called with the number of days (corresponding
    to the desired length of the 'prize strings'), and the
    initial values for the number of consecutive absent days and
    number of total late days.

    >>> _calculate(days=4, absent=0, late=0)
    43
    >>> _calculate(days=30, absent=2, late=0)
    0
    >>> _calculate(days=30, absent=1, late=0)
    98950096
    """

    # if we are absent twice, or late 3 consecutive days,
    # no further prize strings are possible
    if late == 3 or absent == 2:
        return 0

    # if we have no days left, and have not failed any other rules,
    # we have a prize string
    if days == 0:
        return 1

    # No easy solution, so now we need to do the recursive calculation

    # First, check if the combination is already in the cache, and
    # if yes, return the stored value from there since we already
    # know the number of possible prize strings from this point on
    key = (days, absent, late)
    if key in cache:
        return cache[key]

    # now we calculate the three possible ways that can unfold from
    # this point on, depending on our attendance today

    # 1) if we are late (but not absent), the "absent" counter stays as
    # it is, but the "late" counter increases by one
    state_late = _calculate(days - 1, absent, late + 1)

    # 2) if we are absent, the "absent" counter increases by 1, and the
    # "late" counter resets to 0
    state_absent = _calculate(days - 1, absent + 1, 0)

    # 3) if we are on time, this resets the "late" counter and keeps the
    # absent counter
    state_ontime = _calculate(days - 1, absent, 0)

    prizestrings = state_late + state_absent + state_ontime

    cache[key] = prizestrings
    return prizestrings


def solution(days: int = 30) -> int:
    """
    Returns the number of possible prize strings for a particular number
    of days, using a simple recursive function with caching to speed it up.

    >>> solution()
    1918080160
    >>> solution(4)
    43
    """

    return _calculate(days, absent=0, late=0)


if __name__ == "__main__":
    print(solution())