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Project Euler problem148 #8662
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""" | ||
Project Euler Problem 148 : https://projecteuler.net/problem=148 | ||
Author: Sai Teja Manchi | ||
Problem Statement: | ||
We can easily verify that none of the entries in the | ||
first seven rows of Pascal's triangle are divisible by 7: | ||
1 | ||
1 1 | ||
1 2 1 | ||
1 3 3 1 | ||
1 4 6 4 1 | ||
1 5 10 10 5 1 | ||
1 6 15 20 15 6 1 | ||
However, if we check the first one hundred rows, we will find that | ||
only 2361 of the 5050 entries are not divisible by 7. | ||
Find the number of entries which are not divisible by 7 | ||
in the first one billion (109) rows of Pascal's triangle. | ||
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Solution: | ||
We iteratively generate each row in the pascal triangle one-by-one. | ||
Since Pascal's triangle is vertically-symmetric, | ||
We only need to generate half of the values. | ||
We then count the values which are not divisible by 7. | ||
We only store the remainders(when divided by 7) in the list to reduce memory usage. | ||
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Note: In the original problem, we need to calculate for 10^9 rows | ||
but we took 10^4 rows here by default. | ||
""" | ||
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def solution(pascal_row_count: int = 10**4) -> int: | ||
""" | ||
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. use below codes for better efficiency which will solve the timeout issue:-
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. also add docs in |
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To evaluate the solution, use solution() | ||
>>> solution(3) | ||
6 | ||
>>> solution(10) | ||
40 | ||
>>> solution(100) | ||
2361 | ||
""" | ||
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# Initializing pascal row and count | ||
pascal_row = [1, 2] | ||
count = 6 | ||
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# To keep track of length of the pascal row | ||
l = 2 | ||
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for i in range(3, pascal_row_count): | ||
j = 1 | ||
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# Generating the next pascal row | ||
while j < l: | ||
pascal_row[j - 1] = (pascal_row[j - 1] + pascal_row[j]) % 7 | ||
if pascal_row[j - 1] != 0: | ||
count += 2 | ||
j += 1 | ||
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# Adding the middle element for even rows | ||
if i % 2 == 0: | ||
pascal_row[-1] = pascal_row[-1] * 2 | ||
l += 1 | ||
if pascal_row[-1] % 7 != 0: | ||
count += 1 | ||
# Deleting the last element for odd rows since 1 is added at beginning | ||
else: | ||
del pascal_row[-1] | ||
pascal_row.insert(0, 1) | ||
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# Adding 2 to the count for the Additional 1's in the new pascal row | ||
count += 2 | ||
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return count | ||
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if __name__ == "__main__": | ||
print(f"{solution()}") |
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It has a hardcoded value of pascal_row_count which is only 10^4. As a result, the function returns the result for the first 10^4 rows of Pascal's triangle instead of the first billion rows, try this:
def solution(pascal_row_count: int = 10**9) -> int:
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Hey,
I've tried with 10**9 but one of the tests failed due to timeout as it was taking more than 6 hours to run.
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It's not uncommon for brute force solutions to take a long time or run into timeout issues when dealing with very large inputs. In this case, the input value of 10**9 is extremely large, and the current implementation of the "solution" function does not have an efficient algorithm to handle such large inputs.