599. Minimum Index Sum of Two Lists


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:

["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".


Example 2:

["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).



  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.




class Solution {
    vector<string> findRestaurant(vector<string>& list1, vector<string>& list2) {
        vector<string> res;
        unordered_map<string, int> m;
        int mn = INT_MAX, n1 = list1.size(), n2 = list2.size();
        for (int i = 0; i < n1; ++i) m[list1[i]] = i;
        for (int i = 0; i < n2; ++i) {
            if (m.count(list2[i])) {
                int sum = i + m[list2[i]];
                if (sum == mn) res.push_back(list2[i]);
                else if (sum < mn) {
                    mn = sum;
                    res = {list2[i]};
        return res;



Intersection of Two Linked Lists





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