1733. Minimum Number of People to Teach

Problem Description

In a given social network, there are m users and certain friendships between them. A friendship allows two users to communicate only if there's a common language between them. There are n distinct languages, and each user may know one or more languages. The languages array consists of sets, each set containing the languages known by the i-th user. The friendships array consists of pairs, with each pair representing a friendship between two users.

The task is to determine the minimum number of users needed to teach exactly one language so that every friendship in the network has a common language. This will enable all friends to communicate with each other. It's important to note that friendships are not transitive, meaning that mutual friendships do not imply indirect friendships.

Intuition

To solve this problem, we start by identifying all pairs of users within the friendships array that cannot communicate because they do not share a common language. We will store these users in a set to avoid duplication, as teaching a user multiple times is redundant.

After determining the set of users who need to be taught a new language, we will count how many of these users know each language. This is done because the optimal solution would likely involve teaching the most common language among these users, as it will reduce the total number of teachings required.

The key steps in the intuition behind the solution are:

1. Identify the users who cannot communicate with their friends due to language barriers.
2. Among these users, count the occurrence of each known language.
3. Deduct the highest occurrence from the total users who need to be taught to obtain the minimum number of teachings.

The provided Python solution precisely follows this approach, using a helper function check to test if two users can communicate and accumulates the users who need to be taught in a set. It then uses a Counter to tally the languages, and subtracts the most common language's count from the total number of users to teach.

Learn more about Greedy patterns.

Solution Approach

The solution follows a straightforward approach which can be broken down into several steps with corresponding algorithms and data structures.

Firstly, we define a check function that takes two user IDs as inputs and checks whether those two users share at least one language. This function iterates over the language sets of both users (user u and user v), checking for a common language. If a common language is found, it returns True, and if not, False.

Secondly, we create a set s to store all users who cannot communicate with their friends. We iterate through the friendships list and use the check function to determine if a user pair does not have a common language. If not, both users are added to the set s, as they will require teaching a common language to communicate.

Next, after identifying all users who require teaching, we use a Counter from Python's collections module to keep track of the occurrences of each language among these users. The Counter is a dictionary subclass designed for counting hashable objects. Here, it is particularly useful because it allows us to easily increment counts of each language known by the users in set s.

Finally, we identify the language known by the largest number of users who need language teaching by using the max function on the values of the Counter object. We subtract this value from the length of set s, which gives us the minimum number of users who need to be taught a new language. By teaching the most common language among those who can't communicate, we minimize the number of teachings needed.

The solution leverages Python's set and dictionary (via Counter) data structures to efficiently keep track of unique elements and their counts respectively. The use of these data structures, along with effective use of iteration and condition checking, allows for a solution that is relatively easy to understand and efficient in terms of both time and space complexity.

Here's how the implementation looks in reference to the steps above:

1class Solution:
2    def minimumTeachings(
3        self, n: int, languages: List[List[int]], friendships: List[List[int]]
4    ) -> int:
5        # Helper function to check for common language between two users
6        def check(u, v):
7            for x in languages[u - 1]:
8                for y in languages[v - 1]:
9                    if x == y:
10                        return True
11            return False
12
13        # Set to track users who cannot communicate with at least one friend
14        s = set()
15        for u, v in friendships:
16            if not check(u, v):
17                s.add(u)
18                s.add(v)
19
20        # Counter to count the occurrences of each language among the users in set 's'
21        cnt = Counter()
22        for u in s:
23            for l in languages[u - 1]:
24                cnt[l] += 1
25
26        # Subtracting the maximum language occurrence from the number of users to teach
27        return len(s) - max(cnt.values(), default=0)

The solution's time complexity is O(F*L^2) where F is the number of friendships and L is the maximum number of languages a user can know. The space complexity is O(U + L) where U is the number of users who can't communicate with their friends.

Example Walkthrough

Let's walk through a small example to illustrate the solution approach. Suppose we have the following scenario:

• There are n = 3 languages.
• There are m = 4 users with their known languages in the languages array:
  • User 1 knows languages {1, 2}.
  • User 2 knows languages {2, 3}.
  • User 3 knows only language {3}.
  • User 4 knows only language {1}.
• The friendships array indicates friendship pairs:
  • (1, 2): User 1 is friends with User 2.
  • (3, 4): User 3 is friends with User 4.
• We want to enable all friends to communicate with each other by teaching the least number of users a new language.

Now, we apply the solution steps:

1. Identify users who cannot communicate:

   • For friendship (1, 2), User 1 and User 2 share language 2, so they can communicate.
   • For friendship (3, 4), User 3 and User 4 do not share any common language, therefore they cannot communicate.

2. Users who need to learn a new language: User 3 and User 4 should be considered for learning a new language. We add them to the set s = {3, 4}.

3. Count the occurrences of each known language:

   • Users in set s know the following languages:
     • User 3 knows {3}.
     • User 4 knows {1}.
   • Counting these, we get that languages 1 and 3 are known by one user each in need of teaching.

4. Determine the most common language among them:

   • Since both languages 1 and 3 are equally common (each known by one user who can't communicate), we can choose either to teach. Let's say we choose language 1 for maximum coverage.

5. Calculate the minimum number of users to teach:

   • Since we are teaching the most common language (language 1 in our case), and one user (User 4) already knows it, we need to teach only one additional user (User 3).
   • Thus, the result is len(s) - max(cnt.values()) which equals 2 - 1 = 1.
   • We only need to teach 1 user (User 3) a new language to ensure all friends can communicate.

The Python solution identifies these steps programmatically and efficiently performs these calculations, arriving at the optimal solution with minimal teaching required.

Solution Implementation

Time and Space Complexity

Time Complexity

The provided code consists of two major parts - the check for whether each pair of friends speak a common language and the counting of languages among the users who need to be taught.

1. The check function involves two nested loops over the languages spoken by users u and v. If m represents the maximum number of languages any user knows, this function could take up to O(m^2) time in the worst case.

2. The outer loop for iterating over the friendships list, which calls the check function, runs k times where k is the number of friendships. So, this part of the algorithm will take O(k * m^2) time.

3. In the worst case, the set s could include all users, which will, therefore, contain at most 2 * k elements, as each friendship involves two users.

4. The counting of languages again requires an iteration over set s, and for each user in s, iterating over the maximum number of languages they know. In the worst case, this is O(k * m).

5. The final step is finding the maximum value in the counter, which can take O(n) time, where n is the total number of languages.

Combining these elements, the worst-case time complexity of the code is the sum of these complexities: O(k * m^2 + k * m + n).

Space Complexity

1. The check function operates in constant space.

2. The set s can contain up to 2 * k elements, therefore O(k) space complexity.

3. The cnt Counter objects can contain at most n key-value pairs representing the languages, thus O(n) space is needed.

4. Temporary space for iterating and counting - This space is constant.

The combined space complexity of the provided code is the maximum of these, which is O(k + n).

Learn more about how to find time and space complexity quickly using problem constraints.