[LeetCode] 0268. Missing Number

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內容目錄

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?

Python

				
					# time complexity: O(n)
# space complexity: O(1)
from typing import List


class Solution:
    def missingNumber(self, nums: List[int]) -> int:
        numSet = set(nums)
        for i in range(len(numSet) + 1):
            if i not in numSet:
                return i


nums = [9, 6, 4, 2, 3, 5, 7, 0, 1]
print(Solution().missingNumber(nums))

Easy

Python

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