There is a solution which in terms of big-O notation is not that good as binary search, but good enough for the most of modern numbers.

The first instinct you want to go is:

class Solution:
  def mySqrt(self, x: int) -> int:
    i = 0
    step = 1
    while (i+1) * (i+1) <= x:
      i += 1
    return i

This algo is O(X^{1/2}) which is not bad and will work on any modern CPU all the way till 10^14-10^16

But you quickly realise that if your input is, say, 10**16, and you do i = 1, then likely do i=2,3,4,5 is pretty useless.

Instead we can do an incremental jump size too to increase the growth. Once we cross the bar, we slow down the growth until it stops:

class Solution:
  def mySqrt(self, x: int) -> int:
    i = 0
    step = 1
    while (i+step) * (i+step) <= x:
      i += step
      step += 1
    
    while step > 0:
      if (i+step) * (i+step) > x:
        step -= 1
      else:
        i += 1
    return i

This solution is quadratic root of X, and it is not the most perfect still, for 10^1000 the binary search is unbelievably better, but for 10**12 it will do the job in just 1000+~tail of operations, but not exceeding 2000.