(LeetCode) 268. Missing Number

268. Missing Number

• Explore : Interview > Top Interveiw Questions > Easy Collection
• 분류 : Others
• 난이도 : Easy

Problem

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.

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.

Example 4

Input: nums = [0]
Output: 1
Explanation: n = 1 since there is 1 number, so all numbers are in the range [0,1]. 1 is the missing number in the range since it does not appear in nums.

Constraints

• n == nums.length
• 1 <= n <= 10^4
• 0 <= nums[i] <= n
• All the numbers of nums are unique.

Solution

Exapnder

class Solution {
    fun missingNumber(nums: IntArray): Int = nums.size * (nums.size + 1) / 2 - nums.sum()
}

Point of Thinking

• n은 nums.length 이고, nums를 구성하는 모든 element는 유일하다.
• 따라서 nums의 구성 요소의 총합은, 1부터 n까지의 합과 동일해야 한다.
• 주어진 문제는 빠진 숫자 하나를 찾는 것이므로, 1부터 n까지의 합에서 nums의 합계값을 빼면 바로 구해진다.