Introduction

In this blog post, we will explore how to solve the LeetCode problem "1733" 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

On a social network consisting of m users and some friendships between users, two users can communicate with each other if they know a common language. You are given an integer n, an array languages, and an array friendships where: There are n languages numbered 1 through n, languages[i] is the set of languages the i​​​​​​th​​​​ user knows, and friendships[i] = [u​​​​​​i​​​, v​​​​​​i] denotes a friendship between the users u​​​​​​​​​​​i​​​​​ and vi. You can choose one language and teach it to some users so that all friends can communicate with each other. Return the minimum number of users you need to teach. Note that friendships are not transitive, meaning if x is a friend of y and y is a friend of z, this doesn't guarantee that x is a friend of z. Example 1: Input: n = 2, languages = [[1],[2],[1,2]], friendships = [[1,2],[1,3],[2,3]] Output: 1 Explanation: You can either teach user 1 the second language or user 2 the first language. Example 2: Input: n = 3, languages = [[2],[1,3],[1,2],[3]], friendships = [[1,4],[1,2],[3,4],[2,3]] Output: 2 Explanation: Teach the third language to users 1 and 3, yielding two users to teach. Constraints: 2 <= n <= 500 languages.length == m 1 <= m <= 500 1 <= languages[i].length <= n 1 <= languages[i][j] <= n 1 <= u​​​​​​i < v​​​​​​i <= languages.length 1 <= friendships.length <= 500 All tuples (u​​​​​i, v​​​​​​i) are unique languages[i] contains only unique values

Explanation

Here's a breakdown of the solution:

• Identify Non-Communicating Friends: Iterate through the friendships list and check if the corresponding users can communicate (share a language). Store the pairs of friends who cannot communicate.
• Count Language Requirements: For each language, iterate through the non-communicating friend pairs and count how many users in those pairs would need to learn that language for the friends to communicate.

• Find Minimum Users to Teach: Select the language that requires teaching the fewest users.

• Runtime Complexity: O(m * n * k), where m is the number of friendships, n is the number of languages, and k is the maximum size of a user's language list. Storage Complexity: O(m + n), where m is number of friendships and n is number of languages.

Code

    def minimumTeachings(n: int, languages: list[list[int]], friendships: list[list[int]]) -> int:
    """
    Calculates the minimum number of users to teach a language so that all friends can communicate.

    Args:
        n: The number of languages.
        languages: A list of lists, where languages[i] is the set of languages the i-th user knows.
        friendships: A list of lists, where friendships[i] = [u, v] denotes a friendship between users u and v.

    Returns:
        The minimum number of users to teach.
    """

    non_communicating_friends = []
    for u, v in friendships:
        u -= 1  # Adjust to 0-based indexing
        v -= 1
        can_communicate = False
        for lang_u in languages[u]:
            if lang_u in languages[v]:
                can_communicate = True
                break
        if not can_communicate:
            non_communicating_friends.append((u, v))

    min_users_to_teach = float('inf')
    for language in range(1, n + 1):
        users_to_teach = set()
        for u, v in non_communicating_friends:
            if language not in languages[u]:
                users_to_teach.add(u)
            if language not in languages[v]:
                users_to_teach.add(v)
        min_users_to_teach = min(min_users_to_teach, len(users_to_teach))

    return min_users_to_teach