# Minimum Index Sum of Two Lists

`HashMap`, `String`

**Easy**

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.

## Solution

### HashMap - Time: O(n1 + n2), Space - O(n1 \* x)

Note about `clear()`: The time complexity of`ArrayList.clear()`is`O(n)`and of`removeAll`is`O(n^2)`.

```java
class Solution {
    public String[] findRestaurant(String[] list1, String[] list2) {
        ArrayList<String> res = new ArrayList<>();
        HashMap<String, Integer> map = new HashMap<>();
        int minIndexSum = Integer.MAX_VALUE;

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

        for (int j = 0; j < list2.length; j++) {
            if (map.containsKey(list2[j])) {
                int sum = j + map.get(list2[j]);
                if (sum < minIndexSum) {
                    res.clear();
                    res.add(list2[j]);
                    minIndexSum = sum; 
                } else if (sum == minIndexSum) {
                    res.add(list2[j]);
                }
            }
        }
        return res.toArray(new String[res.size()]);
    }
}
```

### HashMap - Time: O(n1 + n2), Space Optimized: O(min(n1, n2) \* x)

Here, `x` refers to the average string length.

Always use the shorter list for building hashmap:

```java
if (list1.length > list2.length) {
    return findRestaurant(list2, list1);
}
```

\--

```java
class Solution {
    public String[] findRestaurant(String[] list1, String[] list2) {
        if (list1.length > list2.length) {
            return findRestaurant(list2, list1);
        }
        ArrayList<String> res = new ArrayList<>();
        HashMap<String, Integer> map = new HashMap<>();
        int minIndexSum = Integer.MAX_VALUE;

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

        for (int j = 0; j < list2.length; j++) {
            if (map.containsKey(list2[j])) {
                int sum = j + map.get(list2[j]);
                if (sum < minIndexSum) {
                    res.clear();
                    res.add(list2[j]);
                    minIndexSum = sum; 
                } else if (sum == minIndexSum) {
                    res.add(list2[j]);
                }
            }
        }
        return res.toArray(new String[res.size()]);
    }
}
```

## Reference

<https://leetcode.com/problems/minimum-index-sum-of-two-lists/solution/>