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Commit 145879b

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mindauglMaximSmolskiypre-commit-ci[bot]
authored
Add solution for the Euler project problem 164. (#12663)
* Add solution for the Euler project problem 164. * Update sol1.py * Update sol1.py * Update sol1.py * Update sol1.py * [pre-commit.ci] auto fixes from pre-commit.com hooks for more information, see https://pre-commit.ci * Update sol1.py --------- Co-authored-by: Maxim Smolskiy <mithridatus@mail.ru> Co-authored-by: pre-commit-ci[bot] <66853113+pre-commit-ci[bot]@users.noreply.github.com>
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‎project_euler/problem_164/__init__.py

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‎project_euler/problem_164/sol1.py

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"""
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Project Euler Problem 164: https://projecteuler.net/problem=164
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Three Consecutive Digital Sum Limit
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How many 20 digit numbers n (without any leading zero) exist such that no three
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consecutive digits of n have a sum greater than 9?
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Brute-force recursive solution with caching of intermediate results.
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"""
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def solve(
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digit: int, prev1: int, prev2: int, sum_max: int, first: bool, cache: dict[str, int]
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) -> int:
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"""
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Solve for remaining 'digit' digits, with previous 'prev1' digit, and
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previous-previous 'prev2' digit, total sum of 'sum_max'.
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Pass around 'cache' to store/reuse intermediate results.
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>>> solve(digit=1, prev1=0, prev2=0, sum_max=9, first=True, cache={})
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9
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>>> solve(digit=1, prev1=0, prev2=0, sum_max=9, first=False, cache={})
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10
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"""
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if digit == 0:
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return 1
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cache_str = f"{digit},{prev1},{prev2}"
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if cache_str in cache:
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return cache[cache_str]
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comb = 0
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for curr in range(sum_max - prev1 - prev2 + 1):
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if first and curr == 0:
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continue
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comb += solve(
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digit=digit - 1,
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prev1=curr,
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prev2=prev1,
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sum_max=sum_max,
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first=False,
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cache=cache,
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)
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cache[cache_str] = comb
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return comb
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def solution(n_digits: int = 20) -> int:
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"""
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Solves the problem for n_digits number of digits.
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>>> solution(2)
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45
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>>> solution(10)
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21838806
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"""
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cache: dict[str, int] = {}
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return solve(digit=n_digits, prev1=0, prev2=0, sum_max=9, first=True, cache=cache)
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if __name__ == "__main__":
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print(f"{solution(10) = }")

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