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

Pre analysis

Will create a map for each array and then compare the two maps to find the difference.

Post analysis

I could have used set instead of map to make the code more readable.

/**
 * @param {number[]} nums1
 * @param {number[]} nums2
 * @return {number[][]}
 */
var findDifference = function (nums1, nums2) {
  let ans1 = new Set(nums1);
  let ans2 = new Set(nums2);

  nums2.forEach((n2) => {
    ans1.delete(n2);
  });

  nums1.forEach((n1) => {
    ans2.delete(n1);
  });

  return [[...ans1], [...ans2]];
};