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Tested Python solution for LeetCode 2215 with 16 pytest cases. Generate a practice environment with lcpy.
lcpy gen -n 2215 # by problem number
lcpy gen -s find_the_difference_of_two_arrays # by problem name
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.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 nums1. Therefore, answer[1] = [4,6].
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] = [].
class Solution:
# Time: O(n + m)
# Space: O(n + m)
def find_difference(self, nums1: list[int], nums2: list[int]) -> list[list[int]]:
set1, set2 = set(nums1), set(nums2)
return [sorted(set1 - set2), sorted(set2 - set1)]
| Time | Space |
|---|---|
| O(n + m) | O(n + m) |