Leetcode problem 66 - Find the Difference of Two Arrays

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] = [].

/**
 * @param {number[]} nums1
 * @param {number[]} nums2
 * @return {number[][]}
 */
var findDifference = function (nums1, nums2) {
  let set1 = new Set(nums1);
  let set2 = new Set(nums2);
  let result = [];
  result[0] = [...set1].filter((num) => !set2.has(num));
  result[1] = [...set2].filter((num) => !set1.has(num));
  return result;
};

The dominant factor in the time complexity is the filtering step, which is O(n) for each set. Since you perform the filtering on two sets independently, the overall time complexity is O(n + n), which simplifies to O(n).