[Project Euler] Fix code style for multiple problems (#3094)

* Fix code style for Project Euler problems:

- 13, 17, 21
- Default args
- Type hints
- File path

* Fix code style for multiple problems

* Made suggested changes

project_euler/problem_13/sol1.py

@@ -1,30 +1,25 @@
 """
+Problem 13: https://projecteuler.net/problem=13
+
 Problem Statement:
 Work out the first ten digits of the sum of the following one-hundred 50-digit
 numbers.
 """
+import os
 
 
-def solution(array):
-    """Returns the first ten digits of the sum of the array elements.
+def solution():
+    """
+    Returns the first ten digits of the sum of the array elements
+    from the file num.txt
 
-    >>> import os
-    >>> sum = 0
-    >>> array = []
-    >>> with open(os.path.dirname(__file__) + "/num.txt","r") as f:
-    ...     for line in f:
-    ...         array.append(int(line))
-    ...
-    >>> solution(array)
+    >>> solution()
     '5537376230'
     """
-    return str(sum(array))[:10]
+    file_path = os.path.join(os.path.dirname(__file__), "num.txt")
+    with open(file_path, "r") as file_hand:
+        return str(sum([int(line) for line in file_hand]))[:10]
 
 
 if __name__ == "__main__":
-    n = int(input().strip())
-
-    array = []
-    for i in range(n):
-        array.append(int(input().strip()))
-    print(solution(array))
+    print(solution())

project_euler/problem_17/sol1.py

@@ -1,6 +1,6 @@
 """
 Number letter counts
-Problem 17
+Problem 17: https://projecteuler.net/problem=17
 
 If the numbers 1 to 5 are written out in words: one, two, three, four, five,
 then there are 3 + 3 + 5 + 4 + 4 = 19 letters used in total.
@@ -16,7 +16,7 @@ usage.
 """
 
 
-def solution(n):
+def solution(n: int = 1000) -> int:
     """Returns the number of letters used to write all numbers from 1 to n.
     where n is lower or equals to 1000.
     >>> solution(1000)

project_euler/problem_21/sol1.py

@@ -1,5 +1,3 @@
-from math import sqrt
-
 """
 Amicable Numbers
 Problem 21
@@ -15,9 +13,10 @@ and 110; therefore d(220) = 284. The proper divisors of 284 are 1, 2, 4, 71 and
 
 Evaluate the sum of all the amicable numbers under 10000.
 """
+from math import sqrt
 
 
-def sum_of_divisors(n):
+def sum_of_divisors(n: int) -> int:
     total = 0
     for i in range(1, int(sqrt(n) + 1)):
         if n % i == 0 and i != sqrt(n):
@@ -27,7 +26,7 @@ def sum_of_divisors(n):
     return total - n
 
 
-def solution(n):
+def solution(n: int = 10000) -> int:
     """Returns the sum of all the amicable numbers under n.
 
     >>> solution(10000)

project_euler/problem_31/sol1.py

@@ -1,6 +1,7 @@
 """
 Coin sums
-Problem 31
+Problem 31: https://projecteuler.net/problem=31
+
 In England the currency is made up of pound, £, and pence, p, and there are
 eight coins in general circulation:
 
@@ -12,39 +13,39 @@ How many different ways can £2 be made using any number of coins?
 """
 
 
-def one_pence():
+def one_pence() -> int:
     return 1
 
 
-def two_pence(x):
+def two_pence(x: int) -> int:
     return 0 if x < 0 else two_pence(x - 2) + one_pence()
 
 
-def five_pence(x):
+def five_pence(x: int) -> int:
     return 0 if x < 0 else five_pence(x - 5) + two_pence(x)
 
 
-def ten_pence(x):
+def ten_pence(x: int) -> int:
     return 0 if x < 0 else ten_pence(x - 10) + five_pence(x)
 
 
-def twenty_pence(x):
+def twenty_pence(x: int) -> int:
     return 0 if x < 0 else twenty_pence(x - 20) + ten_pence(x)
 
 
-def fifty_pence(x):
+def fifty_pence(x: int) -> int:
     return 0 if x < 0 else fifty_pence(x - 50) + twenty_pence(x)
 
 
-def one_pound(x):
+def one_pound(x: int) -> int:
     return 0 if x < 0 else one_pound(x - 100) + fifty_pence(x)
 
 
-def two_pound(x):
+def two_pound(x: int) -> int:
     return 0 if x < 0 else two_pound(x - 200) + one_pound(x)
 
 
-def solution(n):
+def solution(n: int = 200) -> int:
     """Returns the number of different ways can n pence be made using any number of
     coins?
 
@@ -61,4 +62,4 @@ def solution(n):
 
 
 if __name__ == "__main__":
-    print(solution(int(str(input()).strip())))
+    print(solution(int(input().strip())))

project_euler/problem_31/sol2.py

@@ -1,4 +1,7 @@
-"""Coin sums
+"""
+Problem 31: https://projecteuler.net/problem=31
+
+Coin sums
 
 In England the currency is made up of pound, £, and pence, p, and there are
 eight coins in general circulation:
@@ -29,7 +32,7 @@ Example:
 """
 
 
-def solution(pence: int) -> int:
+def solution(pence: int = 200) -> int:
     """Returns the number of different ways to make X pence using any number of coins.
     The solution is based on dynamic programming paradigm in a bottom-up fashion.
 

project_euler/problem_33/sol1.py

@@ -1,5 +1,5 @@
 """
-Problem:
+Problem 33: https://projecteuler.net/problem=33
 
 The fraction 49/98 is a curious fraction, as an inexperienced
 mathematician in attempting to simplify it may incorrectly believe
@@ -14,27 +14,30 @@ and denominator.
 If the product of these four fractions is given in its lowest common
 terms, find the value of the denominator.
 """
+from fractions import Fraction
+from typing import List
 
 
-def isDigitCancelling(num, den):
+def is_digit_cancelling(num: int, den: int) -> bool:
     if num != den:
         if num % 10 == den // 10:
             if (num // 10) / (den % 10) == num / den:
                 return True
+    return False
 
 
-def solve(digit_len: int) -> str:
+def fraction_list(digit_len: int) -> List[str]:
     """
-    >>> solve(2)
-    '16/64 , 19/95 , 26/65 , 49/98'
-    >>> solve(3)
-    '16/64 , 19/95 , 26/65 , 49/98'
-    >>> solve(4)
-    '16/64 , 19/95 , 26/65 , 49/98'
-    >>> solve(0)
-    ''
-    >>> solve(5)
-    '16/64 , 19/95 , 26/65 , 49/98'
+    >>> fraction_list(2)
+    ['16/64', '19/95', '26/65', '49/98']
+    >>> fraction_list(3)
+    ['16/64', '19/95', '26/65', '49/98']
+    >>> fraction_list(4)
+    ['16/64', '19/95', '26/65', '49/98']
+    >>> fraction_list(0)
+    []
+    >>> fraction_list(5)
+    ['16/64', '19/95', '26/65', '49/98']
     """
     solutions = []
     den = 11
@@ -42,14 +45,24 @@ def solve(digit_len: int) -> str:
     for num in range(den, last_digit):
         while den <= 99:
             if (num != den) and (num % 10 == den // 10) and (den % 10 != 0):
-                if isDigitCancelling(num, den):
+                if is_digit_cancelling(num, den):
                     solutions.append(f"{num}/{den}")
             den += 1
         num += 1
         den = 10
-    solutions = " , ".join(solutions)
     return solutions
 
 
+def solution(n: int = 2) -> int:
+    """
+    Return the solution to the problem
+    """
+    result = 1.0
+    for fraction in fraction_list(n):
+        frac = Fraction(fraction)
+        result *= frac.denominator / frac.numerator
+    return int(result)
+
+
 if __name__ == "__main__":
-    print(solve(2))
+    print(solution())

project_euler/problem_43/sol1.py

@@ -1,4 +1,6 @@
 """
+Problem 43: https://projecteuler.net/problem=43
+
 The number, 1406357289, is a 0 to 9 pandigital number because it is made up of
 each of the digits 0 to 9 in some order, but it also has a rather interesting
 sub-string divisibility property.
@@ -38,11 +40,11 @@ def is_substring_divisible(num: tuple) -> bool:
     return True
 
 
-def compute_sum(n: int = 10) -> int:
+def solution(n: int = 10) -> int:
     """
     Returns the sum of all pandigital numbers which pass the
     divisiility tests.
-    >>> compute_sum(10)
+    >>> solution(10)
     16695334890
     """
     list_nums = [
@@ -55,4 +57,4 @@ def compute_sum(n: int = 10) -> int:
 
 
 if __name__ == "__main__":
-    print(f"{compute_sum(10) = }")
+    print(f"{solution() = }")

project_euler/problem_44/sol1.py

@@ -1,4 +1,6 @@
 """
+Problem 44: https://projecteuler.net/problem=44
+
 Pentagonal numbers are generated by the formula, Pn=n(3n−1)/2. The first ten
 pentagonal numbers are:
 1, 5, 12, 22, 35, 51, 70, 92, 117, 145, ...
@@ -24,11 +26,11 @@ def is_pentagonal(n: int) -> bool:
     return ((1 + root) / 6) % 1 == 0
 
 
-def compute_num(limit: int = 5000) -> int:
+def solution(limit: int = 5000) -> int:
     """
     Returns the minimum difference of two pentagonal numbers P1 and P2 such that
     P1 + P2 is pentagonal and P2 - P1 is pentagonal.
-    >>> compute_num(5000)
+    >>> solution(5000)
     5482660
     """
     pentagonal_nums = [(i * (3 * i - 1)) // 2 for i in range(1, limit)]
@@ -42,4 +44,4 @@ def compute_num(limit: int = 5000) -> int:
 
 
 if __name__ == "__main__":
-    print(f"{compute_num() = }")
+    print(f"{solution() = }")

project_euler/problem_46/sol1.py

@@ -1,4 +1,6 @@
 """
+Problem 46: https://projecteuler.net/problem=46
+
 It was proposed by Christian Goldbach that every odd composite number can be
 written as the sum of a prime and twice a square.
 
@@ -84,5 +86,10 @@ def compute_nums(n: int) -> list[int]:
                 return list_nums
 
 
+def solution() -> int:
+    """Return the solution to the problem"""
+    return compute_nums(1)[0]
+
+
 if __name__ == "__main__":
-    print(f"{compute_nums(1) = }")
+    print(f"{solution() = }")

project_euler/problem_551/sol1.py

@@ -167,7 +167,7 @@ def add(digits, k, addend):
         digits.append(digit)
 
 
-def solution(n):
+def solution(n: int = 10 ** 15) -> int:
     """
     returns n-th term of sequence
 
@@ -197,4 +197,4 @@ def solution(n):
 
 
 if __name__ == "__main__":
-    print(solution(10 ** 15))
+    print(f"{solution() = }")

project_euler/problem_63/sol1.py

@@ -11,16 +11,16 @@ Using these conclusions, we will calculate the result.
 """
 
 
-def compute_nums(max_base: int = 10, max_power: int = 22) -> int:
+def solution(max_base: int = 10, max_power: int = 22) -> int:
     """
     Returns the count of all n-digit numbers which are nth power
-    >>> compute_nums(10, 22)
+    >>> solution(10, 22)
     49
-    >>> compute_nums(0, 0)
+    >>> solution(0, 0)
     0
-    >>> compute_nums(1, 1)
+    >>> solution(1, 1)
     0
-    >>> compute_nums(-1, -1)
+    >>> solution(-1, -1)
     0
     """
     bases = range(1, max_base)
@@ -31,4 +31,4 @@ def compute_nums(max_base: int = 10, max_power: int = 22) -> int:
 
 
 if __name__ == "__main__":
-    print(f"{compute_nums(10, 22) = }")
+    print(f"{solution(10, 22) = }")