2215.py

# https://leetcode.com/problems/find-the-difference-of-two-arrays/
"""
Given two 0-indexed integer arrays nums1 and nums2, return a list answer of size 2 where:

answer[0] is a list of all distinct integers in nums1 which are not present in nums2.
answer[1] is a list of all distinct integers in nums2 which are not present in nums1.
Note that the integers in the lists may be returned in any order.



Example 1:

Input: nums1 = [1,2,3], nums2 = [2,4,6]
Output: [[1,3],[4,6]]
Explanation:
For nums1, nums1[1] = 2 is present at index 0 of nums2, whereas nums1[0] = 1 and nums1[2] = 3 are not present in nums2. Therefore, answer[0] = [1,3].
For nums2, nums2[0] = 2 is present at index 1 of nums1, whereas nums2[1] = 4 and nums2[2] = 6 are not present in nums2. Therefore, answer[1] = [4,6].
Example 2:

Input: nums1 = [1,2,3,3], nums2 = [1,1,2,2]
Output: [[3],[]]
Explanation:
For nums1, nums1[2] and nums1[3] are not present in nums2. Since nums1[2] == nums1[3], their value is only included once and answer[0] = [3].
Every integer in nums2 is present in nums1. Therefore, answer[1] = [].


Constraints:

1 <= nums1.length, nums2.length <= 1000
-1000 <= nums1[i], nums2[i] <= 1000
"""

from typing import List


class Solution:
    def findDifference(self, nums1: List[int], nums2: List[int]) -> List[List[int]]:
        table_nums1 = {}
        table_nums2 = {}

        for i in range(len(nums1)):
            table_nums1[nums1[i]] = 0

        for i in range(len(nums2)):
            table_nums2[nums2[i]] = 0

        result = [set(), set()]

        for i in range(len(nums1)):
            if table_nums2.get(nums1[i]) is None:
                result[0].add(nums1[i])

        for i in range(len(nums2)):
            if table_nums1.get(nums2[i]) is None:
                result[1].add(nums2[i])

        return [list(member) for member in result]


print(Solution().findDifference(nums1=[1, 2, 3, 3], nums2=[1, 1, 2]))