Sqrt(x)

Problem Statement

Given a non-negative integer x, return the square root of x rounded down to the nearest integer. The returned integer should be non-negative as well.

You must not use any built-in exponent function or operator.

For example, do not use pow(x, 0.5) in c++ or x ** 0.5 in python.

Leetcode link

Examples

Example 1:

Input: x = 4
Output: 2
Explanation: The square root of 4 is 2, so we return 2.


Example 2:

Input: x = 8
Output: 2
Explanation: The square root of 8 is 2.82842..., and since we round it down to the nearest integer, 2 is returned.

Constraints

• 0 <= x <= 2³¹ - 1

Code

Python3 Code

class Solution(object):
    def mySqrt(self, x):
        """
        :type x: int
        :rtype: int
        """
        return int(x**(0.5))