"""
268. Missing Number
Difficulty: Easy
https://leetcode.com/problems/missing-number/

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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: 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.

Example 2:

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.

Example 3:

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

Output: 8

Explanation:

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?
"""


class Solution:
    def missingNumber(self, nums: List[int]) -> int:
        n = len(nums)
        expected_sum = (n * (n + 1)) // 2
        actual_sum = sum(nums)
        return expected_sum - actual_sum