LeetCode-Sqrt(x)

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

class Solution:

def mySqrt(self, x: int) -> int:

min = 0

max = x

i = 0

if x == 1:

return 1

while i < 1:

mid = (min + max) / 2

res = mid ** 2

res = round(res, 4)

if res == x:

i = 1

break

elif res > x:

max = mid

else:

min = mid

return int(mid)

Thank You so much for reading.