/**
 * 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
 * @see <a href="https://leetcode.com/problems/find-the-difference-of-two-arrays/">LeetCode</a>
 */
fun findDifference(nums1: IntArray, nums2: IntArray): List<List<Int>> {
    val nums1Set = nums1.toHashSet()
    val nums2Set = nums2.toHashSet()

    val nums1Diff = nums1Set.subtract(nums2Set).toList()
    val nums2Diff = nums2Set.subtract(nums1Set).toList()

    return listOf(nums1Diff, nums2Diff)
}

/**
 * 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
 * @see <a href="https://leetcode.com/problems/find-the-difference-of-two-arrays/">LeetCode</a>
 */
fun findDifference(nums1: IntArray, nums2: IntArray): List<List<Int>> {
    val nums1Set = nums1.toHashSet()
    val nums2Set = nums2.toHashSet()

    val nums1Diff = nums1Set.subtract(nums2Set).toList()
    val nums2Diff = nums2Set.subtract(nums1Set).toList()

    return listOf(nums1Diff, nums2Diff)
}
