mirror of
https://github.com/TheAlgorithms/Python.git
synced 2024-11-27 15:01:08 +00:00
bc8df6de31
* [pre-commit.ci] pre-commit autoupdate updates: - [github.com/astral-sh/ruff-pre-commit: v0.2.2 → v0.3.2](https://github.com/astral-sh/ruff-pre-commit/compare/v0.2.2...v0.3.2) - [github.com/pre-commit/mirrors-mypy: v1.8.0 → v1.9.0](https://github.com/pre-commit/mirrors-mypy/compare/v1.8.0...v1.9.0) * [pre-commit.ci] auto fixes from pre-commit.com hooks for more information, see https://pre-commit.ci --------- Co-authored-by: pre-commit-ci[bot] <66853113+pre-commit-ci[bot]@users.noreply.github.com>
105 lines
3.1 KiB
Python
105 lines
3.1 KiB
Python
"""
|
|
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())
|