Minimum Index Sum of Two Lists

• LeetCode: Minimum Index Sum of Two Lists

• Problem:

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.

• Note:

1. The length of both lists will be in the range of [1, 1000].

2. The length of strings in both lists will be in the range of [1, 30].

3. The index is starting from 0 to the list length minus 1.

4. No duplicates in both lists.

• Examples:

Input:
["Shogun", "Tapioca Express", "Burger King", "KFC"]
["Piatti", "The Grill at Torrey Pines", "Hungry Hunter Steakhouse", "Shogun"]
Output: ["Shogun"]
Explanation: The only restaurant they both like is "Shogun".

Input:
["Shogun", "Tapioca Express", "Burger King", "KFC"]
["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).

• Analysis:

Use a hashmap to count the index of one of the input lists.

Use the hashmap to get the minimal index sum of the common string of both strings of lists.

• Code - Java:

class Solution {
    public String[] findRestaurant(String[] list1, String[] list2) {
        Map<String, Integer> indexes = new HashMap<String, Integer>();
        int len1 = list1.length;
        int len2 = list2.length;
        for (int i = 0; i < len1; i++) {
            indexes.put(list1[i], i);
        }
        List<String> list = new ArrayList<String>();
        int sum = len1 + len2 - 2;
        for (int i = 0; i < len2; i++) {
            int count = indexes.getOrDefault(list2[i], -1);
            if (count == -1) {
                continue;
            }
            int temp = count + i;
            if (temp < sum) {
                list.clear();
                sum = temp;
                list.add(list2[i]);
            } else if (temp == sum) {
                list.add(list2[i]);
            }
        }
        int size = list.size();
        String[] results = new String[size];
        for (int i = 0; i < size; i++) {
            results[i] = list.get(i);
        }
        return results;
    }
}

• Code - C++:

class Solution {
public:
    vector<string> findRestaurant(vector<string>& list1, vector<string>& list2) {
        map<string, int> indexes;
        int len1 = list1.size();
        int len2 = list2.size();
        for (int i = 0; i < len1; i++) {
            indexes.insert(pair<string, int>(list1[i], i));
        }
        vector<string> list;
        int sum = len1 + len2 - 2;
        for (int i = 0; i < len2; i++) {
            if (indexes.find(list2[i]) != indexes.end()) {
                int temp = indexes[list2[i]] + i;
                if (temp < sum) {
                    list.clear();
                    sum = temp;
                    list.push_back(list2[i]);
                } else if (temp == sum) {
                    list.push_back(list2[i]);
                }
            }
        }
        return list;
    }
};

• Code - Python:

class Solution(object):
    def findRestaurant(self, list1, list2):
        """
        :type list1: List[str]
        :type list2: List[str]
        :rtype: List[str]
        """
        indexes = {}
        len1, len2 = len(list1), len(list2)
        for i in range(len1):
            indexes[list1[i]] = i
        result = []
        minVal = len1 + len2 - 2
        for i in range(len2):
            if list2[i] in indexes:
                temp = indexes[list2[i]] + i
                if temp < minVal:
                    del result[:]
                    minVal = temp
                    result.append(list2[i])
                elif temp == minVal:
                    result.append(list2[i])
        return result

• Time Complexity: O(N + M), N is the size of one of the input list as A, M is the other as B.

• Space Complexity: O(N), N is the size of the input list as A