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perf: improve Project Euler problem 030 solution 1 (#6267)

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Improve solution (locally 3+ times - from 3+ seconds to ~1 second)

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Maxim Smolskiy 2022-07-25 19:41:12 +03:00 committed by GitHub
parent 7d9ebee75f
commit 90959212e5
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1 changed files with 19 additions and 13 deletions
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project_euler/problem_030/sol1.py
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@ -1,4 +1,4 @@
""" Problem Statement (Digit Fifth Power ): https://projecteuler.net/problem=30
""" Problem Statement (Digit Fifth Powers): https://projecteuler.net/problem=30
Surprisingly there are only three numbers that can be written as the sum of fourth
powers of their digits:
@ -13,26 +13,32 @@ The sum of these numbers is 1634 + 8208 + 9474 = 19316.
Find the sum of all the numbers that can be written as the sum of fifth powers of their
digits.
(9^5)=59,049‬
59049*7=4,13,343 (which is only 6 digit number )
So, number greater than 9,99,999 are rejected
and also 59049*3=1,77,147 (which exceeds the criteria of number being 3 digit)
So, n>999
and hence a bound between (1000,1000000)
9^5 = 59049
59049 * 7 = 413343 (which is only 6 digit number)
So, numbers greater than 999999 are rejected
and also 59049 * 3 = 177147 (which exceeds the criteria of number being 3 digit)
So, number > 999
and hence a number between 1000 and 1000000
"""
def digitsum(s: str) -> int:
DIGITS_FIFTH_POWER = {str(digit): digit**5 for digit in range(10)}
def digits_fifth_powers_sum(number: int) -> int:
"""
>>> all(digitsum(str(i)) == (1 if i == 1 else 0) for i in range(100))
True
>>> digits_fifth_powers_sum(1234)
1300
"""
i = sum(pow(int(c), 5) for c in s)
return i if i == int(s) else 0
return sum(DIGITS_FIFTH_POWER[digit] for digit in str(number))
def solution() -> int:
return sum(digitsum(str(i)) for i in range(1000, 1000000))
return sum(
number
for number in range(1000, 1000000)
if number == digits_fifth_powers_sum(number)
)
if __name__ == "__main__":