/**
 * The firefighter is responding to an emergency where two lists need to be combined into one without any duplicates.
 * Write a function that takes in two lists and returns a new list that contains all the unique elements from both lists.
 * Use list comprehension to solve this problem.
 *
 * @param list1 A list of elements (can be of any type)
 * @param list2 Another list of elements (can be of any type)
 * @return A new list containing all the unique elements from both list1 and list2
 *
 * Example:
 * combineLists(listOf(1, 2, 3, 4), listOf(3, 4, 5, 6)) => [1, 2, 3, 4, 5, 6]
 */
fun combineLists(list1: List<Any>, list2: List<Any>): List<Any> {

    return (list1 + list2).distinct()
}
/**
 * The firefighter is responding to an emergency where two lists need to be combined into one without any duplicates.
 * Write a function that takes in two lists and returns a new list that contains all the unique elements from both lists.
 * Use list comprehension to solve this problem.
 *
 * @param list1 A list of elements (can be of any type)
 * @param list2 Another list of elements (can be of any type)
 * @return A new list containing all the unique elements from both list1 and list2
 *
 * Example:
 * combineLists(listOf(1, 2, 3, 4), listOf(3, 4, 5, 6)) => [1, 2, 3, 4, 5, 6]
 */
fun combineLists(list1: List<Any>, list2: List<Any>): List<Any> {
/**
 * The firefighter is responding to an emergency where two lists need to be combined into one without any duplicates.
 * Write a function that takes in two lists and returns a new list that contains all the unique elements from both lists.
 * Use list comprehension to solve this problem.
 *
 * @param list1 A list of elements (can be of any type)
 * @param list2 Another list of elements (can be of any type)
 * @return A new list containing all the unique elements from both list1 and list2
 *
 * Example:
 * combineLists(listOf(1, 2, 3, 4), listOf(3, 4, 5, 6)) => [1, 2, 3, 4, 5, 6]
 */
fun combineLists(list1: List<Any>, list2: List<Any>): List<Any> {

    return (list1 + list2).distinct()
}