sol1.py

"""
Project Euler Problem 115: https://projecteuler.net/problem=115

NOTE: This is a more difficult version of Problem 114
(https://projecteuler.net/problem=114).

A row measuring n units in length has red blocks
with a minimum length of m units placed on it, such that any two red blocks
(which are allowed to be different lengths) are separated by at least one black square.

Let the fill-count function, F(m, n),
represent the number of ways that a row can be filled.

For example, F(3, 29) = 673135 and F(3, 30) = 1089155.

That is, for m = 3, it can be seen that n = 30 is the smallest value
for which the fill-count function first exceeds one million.

In the same way, for m = 10, it can be verified that
F(10, 56) = 880711 and F(10, 57) = 1148904, so n = 57 is the least value
for which the fill-count function first exceeds one million.

For m = 50, find the least value of n
for which the fill-count function first exceeds one million.
"""

fromitertoolsimportcount


defsolution(min_block_length: int=50) ->int:
"""
 Returns for given minimum block length the least value of n
 for which the fill-count function first exceeds one million

 >>> solution(3)
 30

 >>> solution(10)
 57
 """

fill_count_functions= [1] *min_block_length

fornincount(min_block_length):
fill_count_functions.append(1)

forblock_lengthinrange(min_block_length, n+1):
forblock_startinrange(n-block_length):
fill_count_functions[n] +=fill_count_functions[
n-block_start-block_length-1
 ]

fill_count_functions[n] +=1

iffill_count_functions[n] >1_000_000:
break

returnn


if__name__=="__main__":
print(f"{solution() =}")