[Leet Code] Missing Number

Leetcode: https://leetcode.com/problems/missing-number/submissions/

Problem:

Given an array nums containing n distinct numbers in the range [0, n], return the only number in the range that is missing from the array.
Follow up: Could you implement a solution using only O(1) extra space complexity and O(n) runtime complexity?

Example:

Input: nums = [3,0,1]
Output: 2
Explanation: 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.
Input: nums = [0,1]
Output: 2
Explanation: 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.

Solution:

class Solution(object):
    def missingNumber(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        
        length = len(nums) + 1
        for i in range(length):
            if i not in nums:
                return i

Explanation:

First we get the length of the list, then we add 1 to this list as that will give us all numbers included in the sequence. Then we check if the number at i is in nums, if not, return i.