Add solution for Project Euler problem 74. (TheAlgorithms#3125)

* Added solution for Project Euler problem 74. Fixes: TheAlgorithms#2695

* Added doctest for solution() in project_euler/problem_74/sol1.py

* Update docstrings and 0-padding of directory name. Reference: TheAlgorithms#3256

project_euler/problem_074/sol1.py

@@ -0,0 +1,111 @@
+"""
+Project Euler Problem 74: https://projecteuler.net/problem=74
+
+The number 145 is well known for the property that the sum of the factorial of its
+digits is equal to 145:
+
+1! + 4! + 5! = 1 + 24 + 120 = 145
+
+Perhaps less well known is 169, in that it produces the longest chain of numbers that
+link back to 169; it turns out that there are only three such loops that exist:
+
+169 → 363601 → 1454 → 169
+871 → 45361 → 871
+872 → 45362 → 872
+
+It is not difficult to prove that EVERY starting number will eventually get stuck in
+a loop. For example,
+
+69 → 363600 → 1454 → 169 → 363601 (→ 1454)
+78 → 45360 → 871 → 45361 (→ 871)
+540 → 145 (→ 145)
+
+Starting with 69 produces a chain of five non-repeating terms, but the longest
+non-repeating chain with a starting number below one million is sixty terms.
+
+How many chains, with a starting number below one million, contain exactly sixty
+non-repeating terms?
+"""
+
+
+DIGIT_FACTORIALS = {
+    "0": 1,
+    "1": 1,
+    "2": 2,
+    "3": 6,
+    "4": 24,
+    "5": 120,
+    "6": 720,
+    "7": 5040,
+    "8": 40320,
+    "9": 362880,
+}
+
+CACHE_SUM_DIGIT_FACTORIALS = {145: 145}
+
+CHAIN_LENGTH_CACHE = {
+    145: 0,
+    169: 3,
+    36301: 3,
+    1454: 3,
+    871: 2,
+    45361: 2,
+    872: 2,
+    45361: 2,
+}
+
+
+def sum_digit_factorials(n: int) -> int:
+    """
+    Return the sum of the factorial of the digits of n.
+    >>> sum_digit_factorials(145)
+    145
+    >>> sum_digit_factorials(45361)
+    871
+    >>> sum_digit_factorials(540)
+    145
+    """
+    if n in CACHE_SUM_DIGIT_FACTORIALS:
+        return CACHE_SUM_DIGIT_FACTORIALS[n]
+    ret = sum([DIGIT_FACTORIALS[let] for let in str(n)])
+    CACHE_SUM_DIGIT_FACTORIALS[n] = ret
+    return ret
+
+
+def chain_length(n: int, previous: set = None) -> int:
+    """
+    Calculate the length of the chain of non-repeating terms starting with n.
+    Previous is a set containing the previous member of the chain.
+    >>> chain_length(10101)
+    11
+    >>> chain_length(555)
+    20
+    >>> chain_length(178924)
+    39
+    """
+    previous = previous or set()
+    if n in CHAIN_LENGTH_CACHE:
+        return CHAIN_LENGTH_CACHE[n]
+    next_number = sum_digit_factorials(n)
+    if next_number in previous:
+        CHAIN_LENGTH_CACHE[n] = 0
+        return 0
+    else:
+        previous.add(n)
+        ret = 1 + chain_length(next_number, previous)
+        CHAIN_LENGTH_CACHE[n] = ret
+        return ret
+
+
+def solution(num_terms: int = 60, max_start: int = 1000000) -> int:
+    """
+    Return the number of chains with a starting number below one million which
+    contain exactly n non-repeating terms.
+    >>> solution(10,1000)
+    28
+    """
+    return sum(1 for i in range(1, max_start) if chain_length(i) == num_terms)
+
+
+if __name__ == "__main__":
+    print(f"{solution() = }")