# 69-sqrtx


Try it on <a href='https://leetcode.com/problems/69-sqrtx'>leetcode</a>

## Description
<div class="description">
<div><p>Given a non-negative integer <code>x</code>,&nbsp;compute and return <em>the square root of</em> <code>x</code>.</p>

<p>Since the return type&nbsp;is an integer, the decimal digits are <strong>truncated</strong>, and only <strong>the integer part</strong> of the result&nbsp;is returned.</p>

<p><strong>Note:&nbsp;</strong>You are not allowed to use any built-in exponent function or operator, such as <code>pow(x, 0.5)</code> or&nbsp;<code>x ** 0.5</code>.</p>

<p>&nbsp;</p>
<p><strong>Example 1:</strong></p>

<pre><strong>Input:</strong> x = 4
<strong>Output:</strong> 2
</pre>

<p><strong>Example 2:</strong></p>

<pre><strong>Input:</strong> x = 8
<strong>Output:</strong> 2
<strong>Explanation:</strong> The square root of 8 is 2.82842..., and since the decimal part is truncated, 2 is returned.</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>0 &lt;= x &lt;= 2<sup>31</sup> - 1</code></li>
</ul>
</div>
</div>

## Solution(Python)
```Python
class Solution:
    def mySqrt(self, x: int) -> int:
        if x == 0 or x == 1:
            return x
        lo = 1
        hi = x
        while lo < hi:
            mi = (lo + hi) >> 1

            if mi * mi == x:
                return mi

            if mi * mi > x:
                hi = mi
            else:
                lo = mi + 1
        return lo - 1

```