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2215. Find the Difference of Two Arrays
Description
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
Solutions
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class Solution { public List<List<Integer>> findDifference(int[] nums1, int[] nums2) { Set<Integer> s1 = convert(nums1); Set<Integer> s2 = convert(nums2); List<List<Integer>> ans = new ArrayList<>(); List<Integer> l1 = new ArrayList<>(); List<Integer> l2 = new ArrayList<>(); for (int v : s1) { if (!s2.contains(v)) { l1.add(v); } } for (int v : s2) { if (!s1.contains(v)) { l2.add(v); } } ans.add(l1); ans.add(l2); return ans; } private Set<Integer> convert(int[] nums) { Set<Integer> s = new HashSet<>(); for (int v : nums) { s.add(v); } return s; } }
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class Solution { public: vector<vector<int>> findDifference(vector<int>& nums1, vector<int>& nums2) { unordered_set<int> s1(nums1.begin(), nums1.end()); unordered_set<int> s2(nums2.begin(), nums2.end()); vector<vector<int>> ans(2); for (int v : s1) if (!s2.count(v)) ans[0].push_back(v); for (int v : s2) if (!s1.count(v)) ans[1].push_back(v); return ans; } };
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class Solution: def findDifference(self, nums1: List[int], nums2: List[int]) -> List[List[int]]: s1, s2 = set(nums1), set(nums2) return [list(s1 - s2), list(s2 - s1)]
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func findDifference(nums1 []int, nums2 []int) [][]int { s1, s2 := make(map[int]bool), make(map[int]bool) for _, v := range nums1 { s1[v] = true } for _, v := range nums2 { s2[v] = true } ans := make([][]int, 2) for v := range s1 { if !s2[v] { ans[0] = append(ans[0], v) } } for v := range s2 { if !s1[v] { ans[1] = append(ans[1], v) } } return ans }
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function findDifference(nums1: number[], nums2: number[]): number[][] { return [ [...new Set<number>(nums1.filter(v => !nums2.includes(v)))], [...new Set<number>(nums2.filter(v => !nums1.includes(v)))], ]; }
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/** * @param {number[]} nums1 * @param {number[]} nums2 * @return {number[][]} */ var findDifference = function (nums1, nums2) { let ans1 = new Set(nums1), ans2 = new Set(nums2); for (let num of nums1) { ans2.delete(num); } for (let num of nums2) { ans1.delete(num); } return [Array.from(ans1), Array.from(ans2)]; };
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class Solution { /** * @param Integer[] $nums1 * @param Integer[] $nums2 * @return Integer[][] */ function findDifference($nums1, $nums2) { $rs = [[], []]; $hashtable1 = array_flip(array_unique($nums1)); $hashtable2 = array_flip(array_unique($nums2)); for ($m = 0; $m < count($nums1); $m++) { if (!isset($hashtable2[$nums1[$m]])) { $rs[0][$m] = $nums1[$m]; $hashtable2[$nums1[$m]] = 1; } } for ($n = 0; $n < count($nums2); $n++) { if (!isset($hashtable1[$nums2[$n]])) { $rs[1][$n] = $nums2[$n]; $hashtable1[$nums2[$n]] = 1; } } return $rs; } }
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use std::collections::HashSet; impl Solution { pub fn find_difference(nums1: Vec<i32>, nums2: Vec<i32>) -> Vec<Vec<i32>> { vec![ nums1 .iter() .filter_map(|&v| if nums2.contains(&v) { None } else { Some(v) }) .collect::<HashSet<i32>>() .into_iter() .collect(), nums2 .iter() .filter_map(|&v| if nums1.contains(&v) { None } else { Some(v) }) .collect::<HashSet<i32>>() .into_iter() .collect() ] } }