# LeetCode 645. Set Mismatch

## 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 <= 104`
• `1 <= nums[i] <= 104`

## 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).