0268. Missing Number

Easy

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 <= 10⁴
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:

        # s = set(nums)
        # for n in range(len(nums)+1):
        #     if n not in s:
        #         return n


#         nums.sort()
#         for i in range(len(nums)):
#             if i != nums[i]:
#                 return i

#         return len(nums)


        total = sum(nums)
        expected_sum = len(nums)*(len(nums)+1)//2
        return expected_sum - total

0268. Missing Number

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