Problem Link

Description


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

Solution


Python3

class Solution:
    def findErrorNums(self, nums: List[int]) -> List[int]:
        N = len(nums)
        count = [0] * (N + 1)
        res = [None, None]
 
        for x in nums:
            count[x] += 1
        
        for x in range(1, N + 1):
            if count[x] == 2:
                res[0] = x
            elif count[x] == 0:
                res[1] = x
            
            if res[0] is not None and res[1] is not None:
                return res
        
        return res