Simple solution using set()
Intuitionuisng set() for faster lookupsApproachConvert nums1 and nums2 into sets (nums1_set and nums2_set) to eliminate duplicates
Use sum() to count how many e
# Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
uisng set() for faster lookups
# Approach
Convert nums1 and nums2 into sets (nums1_set and nums2_set) to eliminate duplicates
Use sum() to count how many elements in nums1 exist in nums2_set, and vice versa

# Complexity
- Time complexity: O(n+m)
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
n is the length of nums1 & m is the length of nums2
Iterating through nums1 and nums2 to create nums1_set and nums2_set takes O(n+m) time.
Iterating through nums1 and nums2 to calculate sum by checking each number takes O(n+m) time.

- Space complexity: O(n+m)
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
space used to store nums1_set is O(n) and nums2_set is O(m)
# Code
```python3 []
class Solution:
    def findIntersectionValues(self, nums1: List[int], nums2: List[int]) -> List[int]:
        nums1_set, nums2_set = set(nums1), set(nums2)

        nums1_counter = sum(1 for i in nums1 if i in nums2_set)
        nums2_counter = sum(1 for i in nums2 if i in nums1_set)
        
        return [nums1_counter, nums2_counter]
```
['Python3']