## Description

https://leetcode.com/problems/set-mismatch/

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 <= 10`

^{4}`1 <= nums[i] <= 10`

^{4}

## Explanation

Counter the element occurrences and find out which element occurs twice and which element is missing.

## Python Solution

```
class Solution:
def findErrorNums(self, nums: List[int]) -> List[int]:
results = []
n = len(nums)
counter = {}
for i, num in enumerate(nums):
counter[num] = counter.get(num, 0) + 1
for key, value in counter.items():
if value == 2:
results.append(key)
for i in range(1, n + 1):
if i not in counter:
results.append(i)
return results
```

- Time Complexity: O(N).
- Space Complexity: O(N).