Solution Explanation for Finding the Topic of Each Post

This problem requires processing two tables, Keywords and Posts, to determine the topic(s) associated with each post based on the keywords present in the post content. The solution involves a database query, specifically using MySQL.

Approach

The core idea is to join the Posts and Keywords tables based on whether keywords from Keywords exist (case-insensitively) within the content of the posts from Posts. Then, we group the results by post_id to aggregate the topics associated with each post. Finally, we handle the case where no topics are found for a post.

MySQL Solution Explained

SELECT
    post_id,
    IFNULL(GROUP_CONCAT(DISTINCT topic_id), 'Ambiguous!') AS topic
FROM
    Posts
    LEFT JOIN Keywords ON INSTR(CONCAT(' ', content, ' '), CONCAT(' ', word, ' ')) > 0
GROUP BY post_id;

1. LEFT JOIN Keywords ON INSTR(CONCAT(' ', content, ' '), CONCAT(' ', word, ' ')) > 0: This is the crucial part. A LEFT JOIN ensures that all posts are included in the result, even if they don't have matching keywords. The ON clause uses the INSTR function to check for the presence of each keyword within the post content. Crucially, we add spaces (' ') before and after both content and word. This handles word boundaries effectively, preventing partial matches. For example, it prevents "war" from matching "warning". INSTR returns the starting position of the keyword if found; otherwise, it returns 0. The condition > 0 filters for successful matches.

2. GROUP BY post_id: This groups the results by post_id, so we get one row per post.

3. GROUP_CONCAT(DISTINCT topic_id): This aggregates the topic_id values for each post. The DISTINCT keyword ensures that we don't have duplicate topic IDs if a post contains multiple keywords belonging to the same topic.

4. IFNULL(..., 'Ambiguous!'): This handles the case where no matching keywords are found for a post. IFNULL checks if GROUP_CONCAT returned NULL (meaning no topics were found) and replaces it with 'Ambiguous!' as specified in the problem statement.

Time Complexity Analysis

The time complexity is dominated by the LEFT JOIN operation and the INSTR function calls within it. In the worst case, the INSTR function might have to compare each keyword against every word in the content of every post. The complexity is therefore roughly O(P * K * C), where:

• P is the number of posts.
• K is the number of keywords.
• C is the average number of words in a post content.

However, in practice, database optimizers can significantly improve performance. The actual time depends on the size of the datasets and the specific database implementation. The GROUP BY and GROUP_CONCAT operations are generally efficient in modern database systems.

Other Languages (Conceptual)

While the problem is inherently database-centric, the core logic can be adapted to other programming languages. The basic steps would be:

1. Read the data: Load the Keywords and Posts data into suitable data structures (e.g., dictionaries or lists of dictionaries in Python).
2. Process posts: Iterate through the posts. For each post:
   • Iterate through the keywords.
   • Check for case-insensitive matches using a suitable string function (e.g., lower() and in in Python).
   • Collect the corresponding topic_ids.
3. Handle ambiguity: If no topic_ids are found, assign "Ambiguous!". Otherwise, sort and format the topic_ids as required.
4. Output results: Create the final result table.

This approach would have similar time complexity characteristics as the SQL query, depending on the efficiency of the string matching operations used. The exact implementation details would vary based on the chosen language and data structures.