# leetcode\_599

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: \["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: 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). Note: The length of both lists will be in the range of \[1, 1000]. The length of strings in both lists will be in the range of \[1, 30]. The index is starting from 0 to the list length minus 1. No duplicates in both lists.

## Solutions

1. **hash map**

```cpp
class Solution {
public:
    vector<string> findRestaurant(vector<string>& list1, vector<string>& list2) {
        unordered_map<string, int> m2;
        int i = 0;
        for (auto & s : list2) m2[s] = i++;

        vector<string> res;
        int minsum = INT_MAX;
        for (int i = 0; i < list1.size(); i++) {
            auto & s1 = list1[i];
            if (!m2.count(s1)) continue;
            int sum = i + m2[s1];
            if (sum < minsum) {
                res = {s1}; minsum = sum;
            }
            else if (sum == minsum)
                res.push_back(s1);
        }

        return res;
    }
};
```