Easy
Suppose Andy and Doris want to choose a restaurant for dinner, and they both have a list of favorite restaurants represented by strings.
You need to help them find out their common interest with the least list index sum. If there is a choice tie between answers, output all of them with no order requirement. You could assume there always exists an answer.
Example 1:
Input: list1 = [“Shogun”,”Tapioca Express”,”Burger King”,”KFC”], list2 = [“Piatti”,”The Grill at Torrey Pines”,”Hungry Hunter Steakhouse”,”Shogun”]
Output: [“Shogun”]
Explanation: The only restaurant they both like is “Shogun”.
Example 2:
Input: list1 = [“Shogun”,”Tapioca Express”,”Burger King”,”KFC”], list2 = [“KFC”,”Shogun”,”Burger King”]
Output: [“Shogun”]
Explanation: The restaurant they both like and have the least index sum is “Shogun” with index sum 1 (0+1).
Constraints:
1 <= list1.length, list2.length <= 1000
1 <= list1[i].length, list2[i].length <= 30
list1[i]
and list2[i]
consist of spaces ' '
and English letters.list1
are unique.list2
are unique.class Solution {
fun findRestaurant(list1: Array<String>, list2: Array<String>): Array<String> {
var min = 1000000
val hm: MutableMap<String, Int> = HashMap()
val result: MutableList<String> = ArrayList()
fillMap(list1, hm)
// find min value
for (i in list2.indices) {
if (hm.containsKey(list2[i])) {
val value = hm[list2[i]]!! + i
// a new min value was found
if (value < min) {
min = value
// Clean the arraylist
result.clear()
// add new min value
result.add(list2[i])
} else if (value == min) {
result.add(list2[i])
}
}
}
return result.toTypedArray()
}
fun fillMap(a: Array<String>, hm: MutableMap<String, Int>) {
for (i in a.indices) {
hm[a[i]] = i
}
}
}