2215. Find the Difference of Two Arrays

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].

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) {
  const set1 = new Set(nums1);
  const set2 = new Set(nums2);
 
  const answer = [
    [...set1].filter((num) => !set2.has(num)), // Distinct in nums1 not in nums2
    [...set2].filter((num) => !set1.has(num)), // Distinct in nums2 not in nums1
  ];
 
  return answer;
};

/**
 * @param {number[]} nums1
 * @param {number[]} nums2
 * @return {number[][]}
 */
 
const helper = (nums1, nums2) => {
  const res1 = new Set();
  const res2 = new Set(nums2);
  for (const num of nums1) {
    if (!res2.has(num)) {
      res1.add(num);
    }
  }
  return Array.from(res1);
};
var findDifference = function (nums1, nums2) {
  return [helper(nums1, nums2), helper(nums2, nums1)];
};

/**
 * @param {number[]} nums1
 * @param {number[]} nums2
 * @return {number[][]}
 */
var findDifference = function (nums1, nums2) {
  const num1Set = new Set(nums1);
  const num2Set = new Set(nums2);
 
  const result1 = [...num1Set].filter((num) => !num2Set.has(num));
  const result2 = [...num2Set].filter((num) => !num1Set.has(num));
 
  return [result1, result2];
};

/**
 * @param {number[]} nums1
 * @param {number[]} nums2
 * @return {number[][]}
 */
var findDifference = function (nums1, nums2) {
  let s1 = new Set(nums1);
  let s2 = new Set(nums2);
  let ans = [[], []];
 
  for (let i of s1) {
    if (!s2.has(i)) {
      ans[0].push(i);
    }
  }
 
  for (let i of s2) {
    if (!s1.has(i)) {
      ans[1].push(i);
    }
  }
 
  return ans;
};