Integer Square Root

p
"""
Integer Square Root Algorithm -- An efficient method to calculate the square root of a
non-negative integer 'num' rounded down to the nearest integer. It uses a binary search
approach to find the integer square root without using any built-in exponent functions
or operators.
* https://en.wikipedia.org/wiki/Integer_square_root
* https://docs.python.org/3/library/math.html#math.isqrt
Note:
- This algorithm is designed for non-negative integers only.
- The result is rounded down to the nearest integer.
- The algorithm has a time complexity of O(log(x)).
- Original algorithm idea based on binary search.
"""

def integer_square_root(num: int) -> int:
"""
Returns the integer square root of a non-negative integer num.
Args:
num: A non-negative integer.
Returns:
The integer square root of num.
Raises:
ValueError: If num is not an integer or is negative.
>>> [integer_square_root(i) for i in range(18)]
[0, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 4, 4]
>>> integer_square_root(625)
25
>>> integer_square_root(2_147_483_647)
46340
>>> from math import isqrt
>>> all(integer_square_root(i) == isqrt(i) for i in range(20))
True
>>> integer_square_root(-1)
Traceback (most recent call last):
...
ValueError: num must be non-negative integer
>>> integer_square_root(1.5)
Traceback (most recent call last):
...
ValueError: num must be non-negative integer
>>> integer_square_root("0")
Traceback (most recent call last):
...
ValueError: num must be non-negative integer
"""
if not isinstance(num, int) or num < 0:
raise ValueError("num must be non-negative integer")

if num < 2:
return num

left_bound = 0
right_bound = num // 2

while left_bound <= right_bound:
mid = left_bound + (right_bound - left_bound) // 2
mid_squared = mid * mid
if mid_squared == num:
return mid

if mid_squared < num:
left_bound = mid + 1
else:
right_bound = mid - 1

return right_bound

if __name__ == "__main__":
import doctest

doctest.testmod()