문제

2215. Find the Difference of Two Arrays

User Accepted:9400
User Tried:9851
Total Accepted:9597
Total Submissions:14280
Difficulty:Easy

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

풀이

class Solution:
    def findDifference(self, nums1: List[int], nums2: List[int]) -> List[List[int]]:
        answer = [[],[]]
        
        for i in range(len(nums1)):
            if nums1[i] not in nums2 and nums1[i] not in answer[0] :
                answer[0].append(nums1[i])
        
        for j in range(len(nums2)):
            if nums2[j] not in nums1 and nums2[j] not in answer[1]:
                answer[1].append(nums2[j])
                
        return answer

당장 생각나는대로 풀었더니 시간복잡도가 반복문 안에서 in(배열을 순회함) 을 찾는 O(N^2)이 되어버렸다.

이 방법보다 훨씬 더 간단하게 풀 수 있다.

class Solution:
    def findDifference(self, nums1: List[int], nums2: List[int]) -> List[List[int]]:
        a = set(nums1)
        b = set(nums2)
        return [a - b, b - a]

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