## 1. Problem

2215. Find the Difference of Two Arrays

Given two 0-indexed integer arrays nums1 and nums2, return a list answer of size 2 where:

- answer[0] is a list of all distinct integers in nums1 which are not present in nums2.
- answer[1] is a list of all distinct integers in nums2 which are not present in nums1.

Note that the integers in the lists may be returned in any order.

Example 1:

```
Input: nums1 = [1,2,3], nums2 = [2,4,6]
Output: [[1,3],[4,6]]
Explanation:
For nums1, nums1[1] = 2 is present at index 0 of nums2, whereas nums1[0] = 1 and nums1[2] = 3 are not present in nums2. Therefore, answer[0] = [1,3].
For nums2, nums2[0] = 2 is present at index 1 of nums1, whereas nums2[1] = 4 and nums2[2] = 6 are not present in nums2. Therefore, answer[1] = [4,6].
```

Example 2:

```
Input: nums1 = [1,2,3,3], nums2 = [1,1,2,2]
Output: [[3],[]]
Explanation:
For nums1, nums1[2] and nums1[3] are not present in nums2. Since nums1[2] == nums1[3], their value is only included once and answer[0] = [3].
Every integer in nums2 is present in nums1. Therefore, answer[1] = [].
```

Constraints:

- 1 <= nums1.length, nums2.length <= 1000
- -1000 <= nums1[i], nums2[i] <= 1000

## 2. Solution

I solved this problem like this.

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

- Step
- Using set, we can solve this problem so easy.

```
class Solution:
def findDifference(self, nums1: List[int], nums2: List[int]) -> List[List[int]]:
return [list(set(nums1) - set(nums2)), list(set(nums2) - set(nums1))]
```