# 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.

Constraints:

• 0 <= x <= 231 - 1

# Solution

class Solution(object):
def mySqrt(self, x):
"""
:type x: int
:rtype: int
"""
# right is set to x + 1 to deal with special cases like x = 0, x = 1
left, right = 0, x + 1
while left < right:
mid = left + (right - left) // 2
# Square root means x*x = y
if mid * mid > x:
right = mid
else:
left = mid + 1
# because of our loop we break when we encounter a mid which when squared is more than our target
# therefore our answer is the previous integer
return left - 1