"""

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.
"""

def missingNumber(nums):
    n = len(nums)
    expectedSum = (n*(n+1))/2
    numsSum = sum(nums)

    return int(expectedSum-numsSum)


if __name__ == '__main__':
    x = [3,0,1]

    print("*")
    print(missingNumber(x))
    print("*")
