simple and using dictionary  89.22 % Time and  95.66 % Space
Intuition\nThe intuition behind solving this problem is to iterate through the given list of integers and keep track of the count of occurrences of each element
# Intuition\nThe intuition behind solving this problem is to iterate through the given list of integers and keep track of the count of occurrences of each element. If any element appears more than twice, it indicates that it is not possible to split the list into two parts such that each part contains the same number of occurrences of each integer.\n# Approach\nIn this approach, we use a dictionary to keep track of the count of occurrences of each element in the list. We iterate through the list of integers, and for each element, we check if it exists in the dictionary. If it does, we increment its count; otherwise, we initialize its count to 1. If at any point, the count of any element becomes greater than 2, we return False, indicating that it is not possible to split the list as required. Otherwise, if all elements occur at most twice, we return True.\n# Complexity\n- Time complexity:\nThe time complexity of this approach is O(n), where n is the length of the input list nums. This is because we iterate through the list once, and for each element, the dictionary operations (checking if an element exists and updating its count) are constant time operations.\n- Space complexity:\nThe space complexity is O(n) in the worst case, where n is the number of unique elements in the input list nums. This is because in the worst case, each distinct element in the list will be stored in the dictionary, along with its count. However, if the range of integers in the input list is limited, the space complexity can be considered closer to O(1).\n# Code\n```\nclass Solution:\n    def isPossibleToSplit(self, nums: List[int]) -> bool:\n        a={}\n        for i in nums:\n            if i in a:\n               a[i]+=1\n            else:\n                a[i]=1\n            if a[i]>2:\n                return False\n        return True\n```
['Python3']