""" 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()