69. Sqrt(x)

Problem Link

Description

---

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.

 

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

Solution

---

Python3

class Solution:
    def mySqrt(self, x: int) -> int:
        if x <= 1: return x
        
        def good(v):
            return v*v <= x
        
        left, right = 0, x
        
        while left < right:
            mid = left + (right - left) // 2
            
            if good(mid):
                left = mid + 1
            else:
                right = mid
        
        return left - 1