<p>Given an array <code>nums</code> containing <code>n</code> distinct numbers in the range <code>[0, n]</code>, return <em>the only number in the range that is missing from the array.</em></p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [3,0,1]
<strong>Output:</strong> 2
<strong>Explanation:</strong> n = 3 since there are 3 numbers, so all numbers are in the range [0,3]. 2 is the missing number in the range since it does not appear in nums.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [0,1]
<strong>Output:</strong> 2
<strong>Explanation:</strong> n = 2 since there are 2 numbers, so all numbers are in the range [0,2]. 2 is the missing number in the range since it does not appear in nums.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums = [9,6,4,2,3,5,7,0,1]
<strong>Output:</strong> 8
<strong>Explanation:</strong> n = 9 since there are 9 numbers, so all numbers are in the range [0,9]. 8 is the missing number in the range since it does not appear in nums.
</pre>

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

<ul>
	<li><code>n == nums.length</code></li>
	<li><code>1 &lt;= n &lt;= 10<sup>4</sup></code></li>
	<li><code>0 &lt;= nums[i] &lt;= n</code></li>
	<li>All the numbers of <code>nums</code> are <strong>unique</strong>.</li>
</ul>

<p>&nbsp;</p>
<p><strong>Follow up:</strong> Could you implement a solution using only <code>O(1)</code> extra space complexity and <code>O(n)</code> runtime complexity?</p>

class Solution:
    def missingNumber(self, nums: List[int]) -> int:
        for i in range(len(nums)+1):
            if not i in nums:
                return i
        return []
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