Problem Description

You have a social network with m users where users need a common language to communicate with each other.

Given:

• n: The total number of languages (numbered from 1 to n)
• languages[i]: An array where each element is a list of languages that user i knows (users are 0-indexed in the array but 1-indexed in friendships)
• friendships[i] = [ui, vi]: An array of friendship pairs between users

The problem is asking you to find the minimum number of users you need to teach one single language so that all friend pairs can communicate.

Key points to understand:

• Two users can communicate if they share at least one common language
• You can only choose ONE language to teach to selected users
• The goal is to ensure every friendship pair has at least one common language
• Friendships are not transitive (if x is friends with y, and y is friends with z, x and z are not necessarily friends)

For example, if user 1 knows languages [1, 2] and user 2 knows languages [3, 4], and they are friends, they cannot communicate. You would need to teach either language 1, 2, 3, or 4 to at least one of them so they can communicate.

The solution approach involves:

1. First identifying which friendship pairs cannot currently communicate (no common language)
2. Collecting all users from these "problematic" friendships into a set
3. For each language, counting how many of these problematic users already know it
4. The answer is the total number of problematic users minus the maximum count from step 3 (choosing the language that the most users already know minimizes teaching)

Intuition

The key insight is that we only need to focus on the friendship pairs that currently cannot communicate - these are the "problematic" pairs where the two users share no common language.

Think about it this way: if two friends already share a language, they can communicate, so we don't need to teach them anything. We only need to worry about pairs that can't communicate.

Once we identify all the users involved in these problematic friendships, we need to choose ONE language to teach to some of them. Which language should we choose?

Here's the clever part: if we choose a language that many of these problematic users already know, we'll need to teach fewer people. For example, if we have 10 problematic users and 7 of them already know language #3, then we only need to teach language #3 to the remaining 3 users.

So our strategy becomes:

1. Find all users who are in at least one "cannot communicate" friendship - let's say there are total such users
2. For each language, count how many of these total users already know it
3. Choose the language with the maximum count - let's call this max_count
4. The answer is total - max_count because we need to teach this language to everyone who doesn't already know it

This greedy approach works because teaching the most popular language (among the problematic users) minimizes the number of people we need to teach. It's like choosing the most common denominator - we want to leverage what people already know as much as possible.

Learn more about Greedy patterns.

Solution Approach

Let's walk through the implementation step by step:

Step 1: Identify Problematic Friendships

First, we need a helper function check(u, v) that determines if two users can communicate:

• It iterates through all languages known by user u and user v
• If any language matches, they can communicate (returns True)
• Otherwise, they cannot communicate (returns False)

Step 2: Collect Problematic Users

We create a set s to store all users who are involved in at least one friendship where communication is not possible:

s = set()
for u, v in friendships:
    if not check(u, v):
        s.add(u)
        s.add(v)

Using a set ensures each user is only counted once, even if they're involved in multiple problematic friendships.

Step 3: Count Language Frequencies

For each user in our problematic set s, we count how many users know each language:

cnt = Counter()
for u in s:
    for l in languages[u - 1]:
        cnt[l] += 1

Note: We use u - 1 because users are 1-indexed in friendships but 0-indexed in the languages array.

Step 4: Calculate the Answer

The minimum number of users to teach is:

return len(s) - max(cnt.values(), default=0)

• len(s) is the total number of problematic users
• max(cnt.values()) is the maximum count of any language among these users
• The difference gives us the minimum number of users who need to learn the most popular language
• We use default=0 to handle the edge case where all friendships can already communicate (empty set s)

This greedy approach ensures we teach the fewest possible users by selecting the language that the most problematic users already know.

Example Walkthrough

Let's walk through a concrete example to illustrate the solution approach.

Given:

• n = 3 (languages numbered 1, 2, 3)
• languages = [[1], [2], [1,2], [3]]
  • User 1 knows: language 1
  • User 2 knows: language 2
  • User 3 knows: languages 1 and 2
  • User 4 knows: language 3
• friendships = [[1,2], [2,4], [3,4]]

Step 1: Identify Problematic Friendships

Let's check each friendship pair:

• Friendship [1,2]: User 1 knows {1}, User 2 knows {2} → No common language ❌
• Friendship [2,4]: User 2 knows {2}, User 4 knows {3} → No common language ❌
• Friendship [3,4]: User 3 knows {1,2}, User 4 knows {3} → No common language ❌

All three friendships are problematic!

Step 2: Collect Problematic Users

From the problematic friendships, we collect all involved users:

• From [1,2]: Add users 1 and 2
• From [2,4]: Add users 2 and 4 (user 2 already added)
• From [3,4]: Add users 3 and 4 (user 4 already added)

Problematic users set s = {1, 2, 3, 4} (all 4 users!)

Step 3: Count Language Frequencies

Count how many problematic users know each language:

• Language 1: User 1 knows it, User 3 knows it → count = 2
• Language 2: User 2 knows it, User 3 knows it → count = 2
• Language 3: User 4 knows it → count = 1

Language frequency: {1: 2, 2: 2, 3: 1}

Step 4: Calculate the Answer

• Total problematic users = 4
• Maximum language count = max(2, 2, 1) = 2
• Users to teach = 4 - 2 = 2

Verification:

If we choose language 1 (known by 2 users already):

• Teach language 1 to users 2 and 4
• Now: [1,2] can communicate via language 1 ✓
• Now: [2,4] can communicate via language 1 ✓
• Now: [3,4] can communicate via language 1 ✓

Similarly, if we choose language 2, we'd teach it to users 1 and 4. Either way, we need to teach exactly 2 users!

Solution Implementation

Time and Space Complexity

Time Complexity: O(k × m² + k × m)

Breaking down the analysis:

• The check function has nested loops that iterate through languages of two users, which takes O(m²) in the worst case where m is the maximum number of languages a person can know
• The first main loop iterates through all friendships (k iterations) and calls check for each, resulting in O(k × m²)
• Building set s takes O(k) in the worst case
• The second loop iterates through users in set s (at most 2k users) and their languages (at most m each), contributing O(k × m)
• Finding the maximum value in the counter takes O(n) where n is the total number of languages
• Overall: O(k × m² + k × m + n), which simplifies to O(k × m²) as the dominant term

Space Complexity: O(k + n)

• Set s stores at most 2k users (two per friendship): O(k)
• Counter cnt stores at most n language counts: O(n)
• Total space: O(k + n)

Note: The reference answer states O(m² × k), which appears to be the same as my analysis O(k × m²) with the terms reordered.

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

Common Pitfalls

1. Incorrectly Checking All Languages Instead of Just Problematic Users

A common mistake is to count language frequencies across ALL users in the network, rather than just the users involved in problematic friendships.

Incorrect approach:

# WRONG: Counting languages for all users
language_count = Counter()
for i, user_languages in enumerate(languages):
    for lang in user_languages:
        language_count[lang] += 1

Why it's wrong: We only need to teach languages to users who are in friendships that cannot communicate. Teaching a language to users who can already communicate with all their friends is unnecessary.

Correct approach:

# RIGHT: Only count languages for users in problematic friendships
language_count = Counter()
for user in users_needing_language:  # Only problematic users
    for language_id in languages[user - 1]:
        language_count[language_id] += 1

2. Index Confusion Between User IDs and Array Indices

Users are 1-indexed in the friendships array but the languages array is 0-indexed. Forgetting to convert between these can cause index errors or incorrect results.

Incorrect approach:

# WRONG: Using user ID directly as array index
for user in users_needing_language:
    for language_id in languages[user]:  # Missing the -1 adjustment
        language_count[language_id] += 1

Correct approach:

# RIGHT: Convert 1-indexed user ID to 0-indexed array position
for user in users_needing_language:
    for language_id in languages[user - 1]:  # Proper index adjustment
        language_count[language_id] += 1

3. Trying to Optimize Language Selection Per Friendship Pair

Some might try to find the optimal language for each friendship pair individually, but the problem requires choosing ONE single language to teach globally.

Incorrect approach:

# WRONG: Trying to solve each friendship independently
total_teachings = 0
for u, v in friendships:
    if not can_communicate(u, v):
        # Find best language for this pair only
        best_for_pair = find_best_language(u, v)
        total_teachings += teach_language(best_for_pair, u, v)

Correct approach: The solution must choose one language that works best globally for all problematic friendships, not optimize each friendship individually.

4. Using Set Operations Incorrectly for Language Comparison

When checking if users can communicate, using set operations without proper conversion can lead to errors.

Potentially problematic approach:

# May be inefficient or error-prone with certain inputs
def can_communicate(user1, user2):
    return bool(set(languages[user1-1]) & set(languages[user2-1]))

While this works, it creates unnecessary set objects for every check. The nested loop approach in the solution is clearer and avoids repeated set creation for frequently checked users.