Problem
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
Solution (Java)
class Solution {
public List<List<Integer>> findDifference(int[] nums1, int[] nums2) {
Set<Integer> set1 = createSet(nums1);
Set<Integer> set2 = createSet(nums2);
return Arrays.asList(getMissing(set1, set2), getMissing(set2, set1));
}
private Set<Integer> createSet(int[] array) {
Set<Integer> set = new HashSet<>();
for (int x : array) {
set.add(x);
}
return set;
}
private List<Integer> getMissing(Set<Integer> first, Set<Integer> second) {
List<Integer> list = new ArrayList<>();
for (int x : first) {
if (!second.contains(x)) {
list.add(x);
}
}
return list;
}
}
Explain:
nope.
Complexity:
- Time complexity : O(n).
- Space complexity : O(n).