Easy solution using Counter()
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time 
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity: O(n)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(n)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution:\n    def findIntersectionValues(self, nums1: List[int], nums2: List[int]) -> List[int]:\n        count1 = Counter(nums1)\n        count2 = Counter(nums2)\n        result = [0, 0]\n        for num in count1:\n            if num in count2:\n                result[0] += count1[num]\n                result[1] += count2[num] \n        return result\n```
['Python3']