Optimize perfect cube detection for large numbers with modular check

maths/perfect_cube.py

@@ -2,13 +2,38 @@ 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)
+    True
     >>> perfect_cube(4)
     False
+    >>> perfect_cube(0)
+    True
+    >>> perfect_cube(1)
+    True
+    >>> perfect_cube(-8)  # Negative perfect cube
+    True
+    >>> perfect_cube(-9)  # Negative non-perfect cube
+    False
+    >>> perfect_cube(10**6)  # Large perfect cube (100^3)
+    True
+    >>> perfect_cube(10**6 + 1)  # Large non-perfect cube
+    False
     """
+    # Handle negative numbers
+    if is_negative := n < 0:
+        n = -n
     val = n ** (1 / 3)
-    return (val * val * val) == n
+    # 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)
 
 
 def perfect_cube_binary_search(n: int) -> bool:
@@ -23,6 +48,28 @@ def perfect_cube_binary_search(n: int) -> bool:
     True
     >>> perfect_cube_binary_search(4)
     False
+    >>> perfect_cube_binary_search(0)
+    True
+    >>> perfect_cube_binary_search(1)
+    True
+    >>> perfect_cube_binary_search(-8)  # Negative perfect cube
+    True
+    >>> perfect_cube_binary_search(-9)  # Negative non-perfect cube
+    False
+    >>> perfect_cube_binary_search(10**6)  # Large perfect cube (100^3)
+    True
+    >>> perfect_cube_binary_search(10**6 + 1)  # Large non-perfect cube
+    False
+    >>> perfect_cube_binary_search(10**18)  # Very large perfect cube (10^6)^3
+    True
+    >>> perfect_cube_binary_search(10**18 + 1)  # Very large non-perfect cube
+    False
+    >>> perfect_cube_binary_search(10**100)  # Extremely large number
+    False
+    >>> perfect_cube_binary_search(10**300)  # Extremely large perfect cube (10^100)^3
+    True
+    >>> perfect_cube_binary_search(10**300 + 1)  # Extremely large non-perfect cube
+    False
     >>> perfect_cube_binary_search("a")
     Traceback (most recent call last):
         ...
@@ -31,21 +78,70 @@ def perfect_cube_binary_search(n: int) -> bool:
     Traceback (most recent call last):
         ...
     TypeError: perfect_cube_binary_search() only accepts integers
+    >>> perfect_cube_binary_search(None)
+    Traceback (most recent call last):
+        ...
+    TypeError: perfect_cube_binary_search() only accepts integers
+    >>> perfect_cube_binary_search([])
+    Traceback (most recent call last):
+        ...
+    TypeError: perfect_cube_binary_search() only accepts integers
     """
     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
-    left = 0
-    right = 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:
+        # For very large numbers, use logarithmic approximation
+        # to get a reasonable upper bound
+        log_n = len(str(n))  # Approximate log10(n)
+        approx_root = 10 ** (log_n // 3)
+        left = max(0, approx_root // 10)
+        right = approx_root * 10
+    else:
+        # For smaller numbers, use the standard approach
+        left, right = 0, n // 2 + 1
+
+    # Binary search
     while left <= right:
-        mid = left + (right - left) // 2
-        if mid * mid * mid == n:
+        mid = (left + right) // 2
+        # Avoid computing mid*mid*mid for very large mid values
+        # by checking if mid is too large first
+        if mid > 10**6 and mid * mid > n // mid:
+            right = mid - 1
+            continue
+
+        cube = mid * mid * mid
+        if cube == n:
             return True
-        elif mid * mid * mid < n:
+        elif cube < n:
             left = mid + 1
         else:
             right = mid - 1
+
     return False