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

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

Author: pre-commit-ci[bot]

maths/perfect_cube.py

@@ -1,10 +1,10 @@
 def perfect_cube(n: int) -> bool:
     """
     Check if a number is a perfect cube or not.
-    
+
     Note: This method uses floating point arithmetic which may be
     imprecise for very large numbers.
-    
+
     >>> perfect_cube(27)
     True
     >>> perfect_cube(64)
@@ -30,12 +30,12 @@ def perfect_cube(n: int) -> bool:
         is_negative = True
     else:
         is_negative = False
-        
+
     val = n ** (1 / 3)
     # Round to avoid floating point precision issues
     rounded_val = round(val)
     result = rounded_val * rounded_val * rounded_val == n
-    
+
     # For negative numbers, we need to check if the cube root would be negative
     return result and not (is_negative and rounded_val == 0)
 
@@ -45,7 +45,7 @@ def perfect_cube_binary_search(n: int) -> bool:
     Check if a number is a perfect cube or not using binary search.
     Time complexity : O(Log(n))
     Space complexity: O(1)
-    
+
     >>> perfect_cube_binary_search(27)
     True
     >>> perfect_cube_binary_search(64)
@@ -91,29 +91,29 @@ def perfect_cube_binary_search(n: int) -> bool:
     """
     if not isinstance(n, int):
         raise TypeError("perfect_cube_binary_search() only accepts integers")
-    
+
     # Handle zero and negative numbers
     if n == 0:
         return True
     if n < 0:
         n = -n
-    
+
     # Quick checks to eliminate obvious non-cubes
     # Check last three digits using modulo arithmetic
     # Only 0, 1, 8, 7, 4, 5, 6, 3, 2, 9 can be cubes mod 10
     # But for cubes, the pattern is more complex
     last_digit = n % 10
     if last_digit not in {0, 1, 8, 7, 4, 5, 6, 3, 2, 9}:
         return False
-    
+
     # More refined check: cubes mod 7 can only be 0, 1, 6
     if n % 7 not in {0, 1, 6}:
         return False
-    
+
     # More refined check: cubes mod 9 can only be 0, 1, 8
     if n % 9 not in {0, 1, 8}:
         return False
-    
+
     # Estimate the cube root using logarithms for very large numbers
     # This gives us a much better initial right bound
     if n > 10**18:
@@ -126,7 +126,7 @@ def perfect_cube_binary_search(n: int) -> bool:
     else:
         # For smaller numbers, use the standard approach
         left, right = 0, n // 2 + 1
-    
+
     # Binary search
     while left <= right:
         mid = (left + right) // 2
@@ -135,18 +135,19 @@ def perfect_cube_binary_search(n: int) -> bool:
         if mid > 10**6 and mid * mid > n // mid:
             right = mid - 1
             continue
-            
+
         cube = mid * mid * mid
         if cube == n:
             return True
         elif cube < n:
             left = mid + 1
         else:
             right = mid - 1
-            
+
     return False
 
 
 if __name__ == "__main__":
     import doctest
+
     doctest.testmod()