69. 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.

when x = 8
I expected to the function to return 2, since int(y) rounds down the number, but I got a ZeroDivisionError on line (estimate = (1/2)*(estimate+(x/estimate)). This is an easy leetcode problem #69 sqrt(x)

Here is my code which I ran on leetcode:

def mySqrt(self, x):
    estimate= 1 + ((x-1)//2)
    difference = abs(x-(estimate*estimate))
    tolerance = .005
    while difference  > tolerance:
        estimate = (1/2)*(estimate+(x/estimate))
        difference  = abs((x-(estimate*estimate)))
    return int(estimate)`

Runtime Error
ZeroDivisionError: integer division or modulo by zero
    estimate = (1/2)*(estimate+(x/estimate))
Line 8 in mySqrt (Solution.py)
    ret = Solution().mySqrt(param_1)
Line 29 in _driver (Solution.py)
    _driver()
Line 39 in <module> (Solution.py)