Introduction

In this blog post, we will explore how to solve the LeetCode problem "599" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

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. There is at least a common string between list1 and list2.

Explanation

Here's a breakdown of the solution:

• Create a HashMap for list1: Store each string in list1 as a key in a hashmap, and its index as the value. This allows for O(1) lookup of string indices.
• Iterate through list2: For each string in list2, check if it exists in the hashmap created from list1. If it does, calculate the index sum.

• Maintain Minimum Index Sum and Results: Keep track of the minimum index sum encountered so far. If the current index sum is less than the minimum, update the minimum and reset the results list with the current string. If the current index sum is equal to the minimum, add the current string to the results list.

• Runtime Complexity: O(m + n), where n is the length of list1 and m is the length of list2.

• Storage Complexity: O(n), where n is the length of list1 (for storing the hashmap).

Code

    def findRestaurant(list1: list[str], list2: list[str]) -> list[str]:
    """
    Finds the common strings with the least index sum in two lists of strings.

    Args:
        list1: The first list of strings.
        list2: The second list of strings.

    Returns:
        A list of common strings with the least index sum.
    """

    map1 = {}
    for i, s in enumerate(list1):
        map1[s] = i

    min_index_sum = float('inf')
    result = []

    for j, s in enumerate(list2):
        if s in map1:
            index_sum = map1[s] + j
            if index_sum < min_index_sum:
                min_index_sum = index_sum
                result = [s]
            elif index_sum == min_index_sum:
                result.append(s)

    return result