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 innums1
which are not present innums2
. -
answer[1]
is a list of all distinct integers innums2
which are not present innums1
.
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
SOLUTION:
class Solution:
def findDifference(self, nums1: List[int], nums2: List[int]) -> List[List[int]]:
nums1 = set(nums1)
nums2 = set(nums2)
return [list(nums1 - nums2), list(nums2 - nums1)]
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