645. Set Mismatch

📋 Đề Bài

You have a set of integers s, which originally contains all the numbers from 1 to n. Unfortunately, due to some error, one of the numbers in s got duplicated to another number in the set, which results in repetition of one number and loss of another number.

You are given an integer array nums representing the data status of this set after the error.

Find the number that occurs twice and the number that is missing and return them in the form of an array.

Example 1:

Input: nums = [1,2,2,4]
Output: [2,3]

Example 2:

Input: nums = [1,1]
Output: [1,2]

Constraints:

2 <= nums.length <= 10⁴
1 <= nums[i] <= 10⁴

🧠 Thuật Toán & Kỹ Thuật

Hash Table (Bảng băm)

⏱️ Thời gian O(n)

💾 Không gian O(n)

💻 Lời Giải

Python 0645-set-mismatch.py

class Solution:
    def findErrorNums(self, nums: List[int]) -> List[int]:
        hashMap = Counter(nums)
        n = len(nums)
        s = 0
        ans = []
        
        for key in hashMap:
            s += key
            if hashMap[key] > 1:
                ans.append(key)
                
        ans.append(n * (n + 1) // 2 - s)
        
        return ans