""" 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 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.
"""


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


obj = Solution()
numarr = [3,0,1]
print(obj.missingNumber(numarr))