(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

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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

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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

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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

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

Point of Thinking

  • nnums.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.