LeetCode-in-Java

599. Minimum Index Sum of Two Lists

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.
• All the stings of list1 are unique.
• All the stings of list2 are unique.

Solution

import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;

public class Solution {
    public String[] findRestaurant(String[] list1, String[] list2) {
        int min = 1000000;
        Map<String, Integer> hm = new HashMap<>();
        List<String> result = new ArrayList<>();
        fillMap(list1, hm);
        // find min value
        for (int i = 0; i < list2.length; i++) {
            if (hm.containsKey(list2[i])) {
                int value = hm.get(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.toArray(new String[result.size()]);
    }

    public void fillMap(String[] a, Map<String, Integer> hm) {
        for (int i = 0; i < a.length; i++) {
            hm.put(a[i], i);
        }
    }
}

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