README_EN.md

599. Minimum Index Sum of Two Lists

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Description

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).

Example 3:

Input: list1 = ["Shogun","Tapioca Express","Burger King","KFC"], list2 = ["KFC","Burger King","Tapioca Express","Shogun"]
Output: ["KFC","Burger King","Tapioca Express","Shogun"]

Example 4:

Input: list1 = ["Shogun","Tapioca Express","Burger King","KFC"], list2 = ["KNN","KFC","Burger King","Tapioca Express","Shogun"]
Output: ["KFC","Burger King","Tapioca Express","Shogun"]

Example 5:

Input: list1 = ["KFC"], list2 = ["KFC"]
Output: ["KFC"]

 

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.

Solutions

Python3

class Solution:
    def findRestaurant(self, list1: List[str], list2: List[str]) -> List[str]:
        ans = []
        mp = {v: i for i, v in enumerate(list2)}
        mi = 2000
        for i, v in enumerate(list1):
            if v in mp:
                t = i + mp[v]
                if t < mi:
                    mi = t
                    ans = [v]
                elif t == mi:
                    ans.append(v)
        return ans

Java

class Solution {

    public String[] findRestaurant(String[] list1, String[] list2) {
        Map<String, Integer> mp = new HashMap<>();
        for (int i = 0; i < list2.length; ++i) {
            mp.put(list2[i], i);
        }
        List<String> ans = new ArrayList<>();
        int mi = 2000;
        for (int i = 0; i < list1.length; ++i) {
            if (mp.containsKey(list1[i])) {
                int t = i + mp.get(list1[i]);
                if (t < mi) {
                    ans = new ArrayList<>();
                    ans.add(list1[i]);
                    mi = t;
                } else if (t == mi) {
                    ans.add(list1[i]);
                }
            }
        }
        return ans.toArray(new String[0]);
    }
}

C++

class Solution {
public:
    vector<string> findRestaurant(vector<string>& list1, vector<string>& list2) {
        unordered_map<string, int> mp;
        for (int i = 0; i < list2.size(); ++i) mp[list2[i]] = i;
        int mi = 2000;
        vector<string> ans;
        for (int i = 0; i < list1.size(); ++i)
        {
            if (mp.count(list1[i]))
            {
                int t = i + mp[list1[i]];
                if (t < mi)
                {
                    ans.clear();
                    ans.push_back(list1[i]);
                    mi = t;
                }
                else if (t == mi)
                {
                    ans.push_back(list1[i]);
                }
            }
        }
        return ans;
    }
};

Go

func findRestaurant(list1 []string, list2 []string) []string {
	mp := make(map[string]int)
	for i, v := range list2 {
		mp[v] = i
	}
	mi := 2000
	var ans []string
	for i, v := range list1 {
		if _, ok := mp[v]; ok {
			t := i + mp[v]
			if t < mi {
				ans = []string{v}
				mi = t
			} else if t == mi {
				ans = append(ans, v)
			}
		}
	}
	return ans
}

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