Easy Python Solution
Intuition\nThe code counts the frequency of integers in the input list using the collections.Counter() function. It then checks if an integer occurs only once i
# Intuition\nThe code counts the frequency of integers in the input list using the collections.Counter() function. It then checks if an integer occurs only once in the input list and does not have adjacent integers. The integers satisfying these conditions are added to the output list.\n\n# Approach\nThe approach taken in the code is to first count the frequency of integers in the input list, and then iterate over the list to check for the required conditions. The check involves looking up the frequency of each integer and verifying that it appears only once and that its neighboring integers do not appear in the list. If the integer satisfies these conditions, it is added to the output list.\n\n# Complexity\n- Time complexity:\nO(n)\n\n- Space complexity:\nO(n)\n\n# Code\n```\nclass Solution:\n    def findLonely(self, nums: List[int]) -> List[int]:\n        numCount = collections.Counter(nums)\n        ans = []\n\n        for i in nums:\n            if numCount[i] == 1 and i - 1 not in numCount and i + 1 not in numCount:\n                ans.append(i)\n        \n        return ans\n```
['Python3']