2215. Find the Difference of Two Arrays

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Problem

Given two 0-indexed integer arrays nums1 and nums2, return a list answer of size 2 where:

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:

Solution

/**
 * @param {number[]} nums1
 * @param {number[]} nums2
 * @return {number[][]}
 */
var findDifference = function(nums1, nums2) {
    var map = {};
    for (var i = 0; i < nums1.length; i++) {
        map[nums1[i]] = (map[nums1[i]] || 0) | 1;
    }
    for (var i = 0; i < nums2.length; i++) {
        map[nums2[i]] = (map[nums2[i]] || 0) | 2;
    }
    var res = [new Set(), new Set()];
    for (var i = 0; i < nums1.length; i++) {
        if (!(map[nums1[i]] & 2)) {
            res[0].add(nums1[i]);
        }
    }
    for (var i = 0; i < nums2.length; i++) {
        if (!(map[nums2[i]] & 1)) {
            res[1].add(nums2[i]);
        }
    }
    return res.map(item => Array.from(item));
};

Explain:

nope.

Complexity: