268. Missing Number

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?

 

---

# https://leetcode.com/problems/missing-number/
'''
1. 아이디어 :
    1) 정렬 시킨 후, while문안에 start<=end 투 포인터로 해당값이랑 인덱스랑 일치하는지 확인한다. 맞을 시 답은 오른쪽에 위치,
    반댈로 틀릴 시 답은 왼쪽에 위치. start를 반환
2. 시간복잡도 :
    1) O(logn) + O(n) = O(n)
    - sort * 2중포문 * 투포인터
3. 자료구조 :
    1) 배열

'''

class Solution:
    def missingNumber(self, n: List[int]) -> int:
        n.sort()
        start=0
        end=len(n)-1
        while start<=end:
            mid=(start+end)//2
            if mid==n[mid]:
                start=mid+1
            elif mid!=n[mid]:
                end=mid-1
        return start