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Sqrt(x)
Difficulty: Easy
Category: DSA
Topics: Math, Binary Search
Asked at: Microsoft, Bloomberg, Amazon, Uber, LinkedIn, Google
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.
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**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.
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**Constraints:**
- `0 <= x <= 231 - 1`
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