# 69. Sqrt(x)

Updated: 2023-03-04
1 min read

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LeetCode problem

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.


## First accepted

class Solution:
def mySqrt(self, x: int, div=2) -> int:
s = x // div
s1 = (s + div) // 2
if s1 * s1 > x:
s1 = self.mySqrt(x, s1)
return s1
else:
return s1

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