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

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Author: pre-commit-ci[bot]

maths/aliquot_sum.py

@@ -1,31 +1,35 @@
 from typing import overload, Tuple
 
+
 @overload
 def aliquot_sum(input_num: int, return_factors: bool = False) -> int: ...
 @overload
-def aliquot_sum(input_num: int, return_factors: bool = True) -> Tuple[int, list[int]]: ...
+def aliquot_sum(
+    input_num: int, return_factors: bool = True
+) -> Tuple[int, list[int]]: ...
+
 
 def aliquot_sum(
     input_num: int, return_factors: bool = False
 ) -> int | Tuple[int, list[int]]:
     """
     Calculates the aliquot sum of a positive integer. The aliquot sum is defined as
     the sum of all proper divisors of a number (all divisors except the number itself).
-    
+
     This implementation uses an optimized O(sqrt(n)) algorithm for efficiency.
-    
+
     Args:
         input_num: Positive integer to calculate aliquot sum for
         return_factors: If True, returns tuple (aliquot_sum, sorted_factor_list)
-    
+
     Returns:
         Aliquot sum if return_factors=False
         Tuple (aliquot_sum, sorted_factor_list) if return_factors=True
-    
+
     Raises:
         TypeError: If input is not an integer
         ValueError: If input is not positive
-        
+
     Examples:
         >>> aliquot_sum(15)
         9
@@ -37,41 +41,41 @@ def aliquot_sum(
     # Validate input type - must be integer
     if not isinstance(input_num, int):
         raise TypeError("Input must be an integer")
-    
+
     # Validate input value - must be positive
     if input_num <= 0:
         raise ValueError("Input must be positive integer")
-    
+
     # Special case: 1 has no proper divisors
     if input_num == 1:
         return (0, []) if return_factors else 0
-    
+
     # Initialize factors list with 1 (always a divisor)
-    factors = [1]  
+    factors = [1]
     total = 1  # Start sum with 1
-    
+
     # Calculate square root as optimization boundary
     sqrt_num = int(input_num**0.5)
-    
+
     # Iterate potential divisors from 2 to square root
     for divisor in range(2, sqrt_num + 1):
         # Check if divisor is a factor
         if input_num % divisor == 0:
             # Add divisor to factors list
             factors.append(divisor)
             total += divisor
-            
+
             # Calculate complement (pair factor)
             complement = input_num // divisor
-            
+
             # Avoid duplicate for perfect squares
-            if complement != divisor:  
+            if complement != divisor:
                 factors.append(complement)
                 total += complement
-    
+
     # Sort factors for consistent output
     factors.sort()
-    
+
     # Return based on return_factors flag
     return (total, factors) if return_factors else total
 
@@ -82,16 +86,16 @@ def classify_number(n: int) -> str:
     - Perfect: aliquot sum = number
     - Abundant: aliquot sum > number
     - Deficient: aliquot sum < number
-    
+
     Args:
         n: Positive integer to classify
-        
+
     Returns:
         Classification string ("Perfect", "Abundant", or "Deficient")
-        
+
     Raises:
         ValueError: If input is not positive
-        
+
     Examples:
         >>> classify_number(6)
         'Perfect'
@@ -103,14 +107,14 @@ def classify_number(n: int) -> str:
     # Validate input
     if n <= 0:
         raise ValueError("Input must be positive integer")
-    
+
     # Special case: 1 is always deficient
     if n == 1:
         return "Deficient"
-    
+
     # Calculate aliquot sum (using only the integer version)
     s = aliquot_sum(n)  # Default returns only integer
-    
+
     # Determine classification
     if s == n:
         return "Perfect"
@@ -119,15 +123,15 @@ def classify_number(n: int) -> str:
 
 if __name__ == "__main__":
     import doctest
-    
+
     # Run embedded doctests for verification
     doctest.testmod()
-    
+
     # Additional demonstration examples
     print("Aliquot sum of 28:", aliquot_sum(28))  # Perfect number
     print("Factors of 28:", aliquot_sum(28, return_factors=True)[1])
     print("Classification of 28:", classify_number(28))
-    
+
     # Large number performance test with targeted exception handling
     try:
         print("\nCalculating aliquot sum for 10^9...")