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Leetcode 599. Minimum Index Sum of Two Lists Sure! Let’s break down the problem-solving process for LeetCode problem 599, “Minimum Index Sum of Two Lists.”

Problem Statement:

Given two lists list1 and list2 of strings, find the common strings with the least index sum. If there is a tie, output all the strings with the minimum index sum.

Clarifying Questions:

  1. Are there any constraints on the length of the lists?
  2. Are there any constraints regarding the contents of the lists (e.g., no duplicate strings within a single list)?
  3. Should the output order matter if there are multiple strings with the same minimum index sum?

Presuming the following:

Strategy:

  1. Create a map to store the indices of the elements in list1.
  2. Iterate through list2 and for each string, check if it exists in list1.
  3. Calculate the index sum for common strings.
  4. Track the minimum index sum and update a list of resulting strings when you find a smaller or equal index sum with a common string.
  5. Return the list of common strings with the smallest index sum.

Code:

import java.util.*;

public class Solution {
    public String[] findRestaurant(String[] list1, String[] list2) {
        // Step 1: Create a map for list1 to store the value and its index
        Map<String, Integer> map = new HashMap<>();
        for (int i = 0; i < list1.length; i++) {
            map.put(list1[i], i);
        }
        
        // Step 2: Initialize the minimum index sum to a large value
        int minIndexSum = Integer.MAX_VALUE;
        List<String> result = new ArrayList<>();

        // Step 3: Traverse list2 and find common strings with minimum index sum
        for (int j = 0; j < list2.length; j++) {
            if (map.containsKey(list2[j])) {
                int indexSum = j + map.get(list2[j]);
                if (indexSum < minIndexSum) {
                    result.clear();
                    result.add(list2[j]);
                    minIndexSum = indexSum;
                } else if (indexSum == minIndexSum) {
                    result.add(list2[j]);
                }
            }
        }
        
        return result.toArray(new String[result.size()]);
    }

    public static void main(String[] args) {
        Solution sol = new Solution();
        String[] list1 = {"Shogun", "Tapioca Express", "Burger King", "KFC"};
        String[] list2 = {"Piatti", "The Grill at Torrey Pines", "Hungry Hunter Steakhouse", "Shogun"};
        System.out.println(Arrays.toString(sol.findRestaurant(list1, list2))); // Output: ["Shogun"]
    }
}

Time Complexity:

The time complexity for this solution is (O(n + m)), where (n) is the length of list1 and (m) is the length of list2. This is because we traverse both lists at most once:

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