Introduction

In this blog post, we will explore how to solve the LeetCode problem "1436" 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

You are given the array paths, where paths[i] = [cityAi, cityBi] means there exists a direct path going from cityAi to cityBi. Return the destination city, that is, the city without any path outgoing to another city. It is guaranteed that the graph of paths forms a line without any loop, therefore, there will be exactly one destination city. Example 1: Input: paths = [["London","New York"],["New York","Lima"],["Lima","Sao Paulo"]] Output: "Sao Paulo" Explanation: Starting at "London" city you will reach "Sao Paulo" city which is the destination city. Your trip consist of: "London" -> "New York" -> "Lima" -> "Sao Paulo". Example 2: Input: paths = [["B","C"],["D","B"],["C","A"]] Output: "A" Explanation: All possible trips are: "D" -> "B" -> "C" -> "A". "B" -> "C" -> "A". "C" -> "A". "A". Clearly the destination city is "A". Example 3: Input: paths = [["A","Z"]] Output: "Z" Constraints: 1 <= paths.length <= 100 paths[i].length == 2 1 <= cityAi.length, cityBi.length <= 10 cityAi != cityBi All strings consist of lowercase and uppercase English letters and the space character.

Explanation

Here's an efficient solution to find the destination city in the given path array:

• Identify Starting Cities: Create a set of all the starting cities (cityA).

• Identify Destination Cities: Iterate through the paths and return the cityB that is not present in the set of starting cities.

• Runtime Complexity: O(n), where n is the number of paths.

• Storage Complexity: O(n), where n is the number of paths (due to the set).

Code

    def destCity(paths: list[list[str]]) -> str:
    """
    Finds the destination city in a list of paths.

    Args:
        paths: A list of paths, where each path is a list of two cities [cityA, cityB],
               representing a direct path from cityA to cityB.

    Returns:
        The destination city, which is the city without any outgoing paths.
    """

    starting_cities = set()
    for path in paths:
        starting_cities.add(path[0])  # Add the starting city to the set

    for path in paths:
        if path[1] not in starting_cities:  # Check if the destination city is not a starting city
            return path[1]

    return ""  # Should not reach here according to the problem description, but added for safety.