# 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 theircommon interestwith theleast 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)`.

``````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();
minIndexSum = sum;
} else if (sum == minIndexSum) {
}
}
}
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:

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

--

``````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();
minIndexSum = sum;
} else if (sum == minIndexSum) {
}
}
}
return res.toArray(new String[res.size()]);
}
}
``````

## Reference

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