[pre-commit.ci] auto fixes from pre-commit.com hooks

for more information, see https://pre-commit.ci

Author: pre-commit-ci[bot]

project_euler/problem_095/sol1.py

@@ -8,27 +8,31 @@
 In this implementation, we aim to identify all amicable chains and find the one with the maximum length, while also returning the smallest member of that chain.
 """
 
+
 def sum_of_proper_divisors(n):
     """Calculate the sum of proper divisors of n."""
     if n < 2:
         return 0  # Proper divisors of 0 and 1 are none.
     total = 1  # Start with 1, since it is a proper divisor of any n > 1
     sqrt_n = int(n**0.5)  # Calculate the integer square root of n.
-    
+
     # Loop through possible divisors from 2 to the square root of n
     for i in range(2, sqrt_n + 1):
         if n % i == 0:  # Check if i is a divisor of n
             total += i  # Add the divisor
             if i != n // i:  # Avoid adding the square root twice
                 total += n // i  # Add the corresponding divisor (n/i)
-    
+
     return total
 
+
 def find_longest_amicable_chain(limit):
     """Find the smallest member of the longest amicable chain under a given limit."""
     sum_divisors = {}  # Dictionary to store the sum of proper divisors for each number
     for i in range(1, limit + 1):
-        sum_divisors[i] = sum_of_proper_divisors(i)  # Calculate and store sum of proper divisors
+        sum_divisors[i] = sum_of_proper_divisors(
+            i
+        )  # Calculate and store sum of proper divisors
 
     longest_chain = []  # To store the longest amicable chain found
     seen = {}  # Dictionary to track numbers already processed
@@ -43,7 +47,9 @@ def find_longest_amicable_chain(limit):
         while current <= limit and current not in chain:
             chain.append(current)  # Add the current number to the chain
             seen[current] = True  # Mark this number as seen
-            current = sum_divisors.get(current, 0)  # Move to the next number in the chain
+            current = sum_divisors.get(
+                current, 0
+            )  # Move to the next number in the chain
 
         # Check if we form a cycle and validate the chain
         if current in chain and current != start:
@@ -54,12 +60,16 @@ def find_longest_amicable_chain(limit):
                 if len(chain) > len(longest_chain):
                     longest_chain = chain  # Update longest chain if this one is longer
 
-    return min(longest_chain) if longest_chain else None  # Return the smallest member of the longest chain
+    return (
+        min(longest_chain) if longest_chain else None
+    )  # Return the smallest member of the longest chain
+
 
 def solution():
     """Return the smallest member of the longest amicable chain under one million."""
     return find_longest_amicable_chain(10**6)
 
+
 if __name__ == "__main__":
     smallest_member = solution()  # Call the solution function
     print(smallest_member)  # Output the smallest member of the longest amicable chain