## Question

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.*

**Example 1:**

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

**Example 2:**

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

**Example 3:**

Input:nums = [9,6,4,2,3,5,7,0,1]Output:8Explanation: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.

**Constraints:**

`n == nums.length`

`1 <= n <= 10`

^{4}`0 <= nums[i] <= n`

- All the numbers of
`nums`

are**unique**.

**Follow up:** Could you implement a solution using only `O(1)`

extra space complexity and `O(n)`

runtime complexity?

## Python Solution

class Solution: def missingNumber(self, nums: List[int]) -> int: n = max(nums) s1 = sum(nums) s2 = n*(n+1)//2 if 0 not in nums: return 0 elif s1==s2: return n+1 else: return (s2 - s1)