[Python3] - 1 LINE Solution || Simple Explanation
Idea\n\n\n##### To solve this question easily, we can take advantage of one of the specifications:\n\nnums[i] is a non-empty array of distinct positive integers
# **Idea**\n<br>\n\n##### To solve this question easily, we can take advantage of one of the specifications:\n\n*nums[i] is a non-empty array of **distinct** positive integers*\n\n<br>\n\n##### Since the integers in each sub-list of nums are distinct (i.e. unique), we can count the number of occurences of every integer in nums.\n<br>\n\n##### Suppose it happens that the number of occurences for some integer x is equal to the length of nums (the number of sub-lists).\n##### It would mean, that x must have appeared in every single sub-list of nums (because it could appear in each of the sub-lists at most 1 time).\n\n<br>\n\n##### Notice, that the result is a list of integers like x, so we just need to find all of them.\n\n<br>\n<br>\n\n# **Code**\n<br>\n\n```\n\nclass Solution:\n    def intersection(self, A: List[List[int]]) -> List[int]:\n        return sorted([k for k,v in Counter([x for l in A for x in l]).items() if v==len(A)])\n\t\t\n```\n\n<br>\n<br>\n\n# **Breakdown**\n\n##### Here\'s a breakdown of each operation that\'s happening in the line above:\n\n<br>\n\n```\n\nclass Solution:\n    def intersection(self, A: List[List[int]]) -> List[int]:\n        # Flattening the list A\n        # For example, if A = [[3, 5, 1], [2, 3, 1]] then flat_list = [3, 5, 1, 2, 3, 1]\n        flat_list = [x for lst in A for x in lst]\n\n        # Counter - a container from the collections module\n        # Here, it is pretty much a dictionary, where:\n        # keys - the integers from flat_list\n        # values - the number of occurences of the keys\n\n        # counts from the above flat_list would be equal to {3: 2, 5: 1, 1: 2, 2: 1}\n        counts = Counter(flat_list)\n\n        # dict.items() is a method that returns a list of pairs that look like (key, value)\n        # Here, items would be equal to [(3, 2), (5, 1), (1, 2), (2, 1)]\n        items = counts.items()\n\n        # result array created by taking every key from items array\n        # which has the value (number of key\'s occurences) equal to the length of A (here, 2)\n        # The result would be equal to [3, 1]\n        result = [key for (key, value) in items if value == len(A)]\n\n        # Sort the result array and return it\n        # The result is [1, 3]\n        return sorted(result)\n\t\t\n```\n\n<br>\n\n---\n\n<br>\n\n\n### Thanks for reading, and have an amazing day! \uFF3C(\uFF3E\u25BD\uFF3E)\uFF0F\n\n<br>
['Python', 'Python3']