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* Project Euler problem 191 solution * Add type hints and reference links * Address requested changes - update documentation - split out helper function but mark it with an underscore - remove redundant comments or make them more explicit/helpful * Address requested changes
106 lines
3.1 KiB
Python
106 lines
3.1 KiB
Python
"""
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Prize Strings
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Problem 191
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A particular school offers cash rewards to children with good attendance and
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punctuality. If they are absent for three consecutive days or late on more
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than one occasion then they forfeit their prize.
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During an n-day period a trinary string is formed for each child consisting
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of L's (late), O's (on time), and A's (absent).
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Although there are eighty-one trinary strings for a 4-day period that can be
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formed, exactly forty-three strings would lead to a prize:
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OOOO OOOA OOOL OOAO OOAA OOAL OOLO OOLA OAOO OAOA
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OAOL OAAO OAAL OALO OALA OLOO OLOA OLAO OLAA AOOO
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AOOA AOOL AOAO AOAA AOAL AOLO AOLA AAOO AAOA AAOL
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AALO AALA ALOO ALOA ALAO ALAA LOOO LOOA LOAO LOAA
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LAOO LAOA LAAO
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How many "prize" strings exist over a 30-day period?
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References:
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- The original Project Euler project page:
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https://projecteuler.net/problem=191
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"""
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cache = {}
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def _calculate(days: int, absent: int, late: int) -> int:
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"""
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A small helper function for the recursion, mainly to have
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a clean interface for the solution() function below.
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It should get called with the number of days (corresponding
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to the desired length of the 'prize strings'), and the
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initial values for the number of consecutive absent days and
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number of total late days.
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>>> _calculate(days=4, absent=0, late=0)
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43
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>>> _calculate(days=30, absent=2, late=0)
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0
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>>> _calculate(days=30, absent=1, late=0)
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98950096
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"""
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# if we are absent twice, or late 3 consecutive days,
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# no further prize strings are possible
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if late == 3 or absent == 2:
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return 0
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# if we have no days left, and have not failed any other rules,
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# we have a prize string
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if days == 0:
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return 1
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# No easy solution, so now we need to do the recursive calculation
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# First, check if the combination is already in the cache, and
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# if yes, return the stored value from there since we already
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# know the number of possible prize strings from this point on
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key = (days, absent, late)
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if key in cache:
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return cache[key]
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# now we calculate the three possible ways that can unfold from
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# this point on, depending on our attendance today
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# 1) if we are late (but not absent), the "absent" counter stays as
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# it is, but the "late" counter increases by one
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state_late = _calculate(days - 1, absent, late + 1)
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# 2) if we are absent, the "absent" counter increases by 1, and the
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# "late" counter resets to 0
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state_absent = _calculate(days - 1, absent + 1, 0)
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# 3) if we are on time, this resets the "late" counter and keeps the
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# absent counter
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state_ontime = _calculate(days - 1, absent, 0)
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prizestrings = state_late + state_absent + state_ontime
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cache[key] = prizestrings
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return prizestrings
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def solution(days: int = 30) -> int:
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"""
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Returns the number of possible prize strings for a particular number
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of days, using a simple recursive function with caching to speed it up.
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>>> solution()
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1918080160
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>>> solution(4)
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43
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"""
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return _calculate(days, absent=0, late=0)
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if __name__ == "__main__":
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print(solution())
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