# 599 Minimum Index Sum of Two Lists

## 599. Minimum Index Sum of Two Lists

## 1. Question

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:**

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.

## 2. Implementation

**(1) Hash Table**

```java
class Solution {
    public String[] findRestaurant(String[] list1, String[] list2) {
        Map<String, Integer> map = new HashMap<>();

        for (int i = 0; i < list1.length; i++) {
            map.put(list1[i], i);
        }

        int minSum = Integer.MAX_VALUE, sum = 0;
        List<String> list = new ArrayList<>();

        for (int j = 0; j < list2.length; j++) {
            if (map.containsKey(list2[j])) {
                sum = j + map.get(list2[j]);

                if (sum < minSum) {
                    list.clear();
                    list.add(list2[j]);
                    minSum = sum;
                }
                else if (sum == minSum) {
                    list.add(list2[j]);
                }
            }
        }
        String[] res = new String[list.size()];
        int index = 0;

        for (String s : list) {
            res[index++] = s;
        }
        return res;
    }
}
```

## 3. Time & Space Complexity

**Hash Table:** 时间复杂度O(m + n)), m为list1的长度, n为list2的长度，空间复杂度O(m)
