# Minimum Index Sum Of Two Lists Problem

## Description

LeetCode Problem 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:

``````1
2
3
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:

``````1
2
3
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:

``````1
2
Input: list1 = ["Shogun","Tapioca Express","Burger King","KFC"], list2 = ["KFC","Burger King","Tapioca Express","Shogun"]
Output: ["KFC","Burger King","Tapioca Express","Shogun"]
``````

Example 4:

``````1
2
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:

``````1
2
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 list2are unique.

## Sample C++ Code

``````1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
class Solution {
public:
vector<string> findRestaurant(vector<string>& list1, vector<string>& list2) {
vector<string> res;
unordered_map<string,int> m;
int min = INT_MAX;
for (int i = 0; i < list1.size(); i++)
m[list1[i]] = i;
for (int i = 0; i < list2.size(); i++)
if (m.count(list2[i]) != 0) {
if (m[list2[i]] + i < min) {
min = m[list2[i]] + i;
res.clear();
res.push_back(list2[i]);
}
else if(m[list2[i]] + i == min) {
res.push_back(list2[i]);
}
}
return res;
}
};
``````