Problem Statement:

Given two arrays of strings list1 and list2, find the common strings with the least index sum.

A common string is a string that appeared in both list1 and list2.

A common string with the least index sum is a common string such that if it appeared at list1[i] and list2[j] then i + j should be the minimum value among all the other common strings.

Return all the common strings with the least index sum. Return the answer in any order.

Example 1:

Input: list1 = ["Shogun","Tapioca Express","Burger King","KFC"], list2 = ["Piatti","The Grill at Torrey Pines","Hungry Hunter Steakhouse","Shogun"]
Output: ["Shogun"]
Explanation: The only common string is "Shogun".

Example 2:

Input: list1 = ["Shogun","Tapioca Express","Burger King","KFC"], list2 = ["KFC","Shogun","Burger King"]
Output: ["Shogun"]
Explanation: The common string with the least index sum is "Shogun" with index sum = (0 + 1) = 1.

Example 3:

Input: list1 = ["happy","sad","good"], list2 = ["sad","happy","good"]
Output: ["sad","happy"]
Explanation: There are three common strings:
"happy" with index sum = (0 + 1) = 1.
"sad" with index sum = (1 + 0) = 1.
"good" with index sum = (2 + 2) = 4.
The strings with the least index sum are "sad" and "happy".

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 strings of list1 are unique.
• All the strings of list2 are unique.

Solution:
Algorithm:

1. Create a list to store the result.
2. Create a variable to store the minimum index sum.
3. Traverse the list1 and list2.
4. If the element of list1 is equal to the element of list2, then check if the index sum is less than the minimum index sum.
5. If the index sum is less than the minimum index sum, then clear the list and add the element to the list.
6. If the index sum is equal to the minimum index sum, then add the element to the list.
7. Return the list as an array

Code:

public class Solution {
    public String[] findRestaurant(String[] list1, String[] list2) {
        int min = Integer.MAX_VALUE;
        List<String> res = new ArrayList<>();
        for (int i = 0; i < list1.length; i++) {
            for (int j = 0; j < list2.length; j++) {
                if (list1[i].equals(list2[j])) {
                    if (i + j < min) {
                        res.clear();
                        res.add(list1[i]);
                        min = i + j;
                    } else if (i + j == min) {
                        res.add(list1[i]);
                    }
                }
            }
        }
        return res.toArray(new String[res.size()]);
    }

}

Time Complexity:
O(n*m) where n is the length of the list1 and m is the length of list2.

Space Complexity:
O(n) where n is the length of the list1 and list2.