Leetcode with Python -> Sort

349. Intersection of Two Arrays

Given two arrays, write a function to compute their intersection.

Example:
Given nums1 = [1, 2, 2, 1], nums2 = [2, 2], return [2].

Note:

• Each element in the result must be unique.
• The result can be in any order.

class Solution(object):
    def intersection(self, nums1, nums2):
        """
        :type nums1: List[int]
        :type nums2: List[int]
        :rtype: List[int]
        """
        return list(set(nums1)&set(nums2))
    # for two set, set1&set2 makes a new set containing the intersection of these two sets,likewise, | makes the union set, and ^ makes the union set with the exception of the intersection elements
    # this solution is clear but not fast

 350. Intersection of Two Arrays II

Given two arrays, write a function to compute their intersection.

Example:
Given nums1 = [1, 2, 2, 1], nums2 = [2, 2], return [2, 2].

Note:

• Each element in the result should appear as many times as it shows in both arrays.
• The result can be in any order.

Follow up:

• What if the given array is already sorted? How would you optimize your algorithm?
• What if nums1's size is small compared to nums2's size? Which algorithm is better?
• What if elements of nums2 are stored on disk, and the memory is limited such that you cannot load all elements into the memory at once?

class Solution(object):
    def intersect(self, nums1, nums2):
        """
        :type nums1: List[int]
        :type nums2: List[int]
        :rtype: List[int]
        """
        from collections import Counter
        c1=Counter(nums1)
        c2=Counter(nums2)
        return sum([[num]*min(c1[num],c2[num]) for num in c1&c2],[])
    
    # sum([[1,2],[3,4]],[]) makes [1,2,3,4], why???
    # Counter(some_list) makes a diction, whose key is the elements of the list and the value is the time it appears in the list
    # for two counter, c1&c2 makes the intersection of their keys and the value is always 1