# 69. Sqrt(x)

# 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 <= 2`

^{31}- 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
```