Solution Explanation for LeetCode 1811: Find Interview Candidates

This problem requires querying two tables, Contests and Users, to identify interview candidates based on their performance in LeetCode contests. The solution involves several steps:

1. Data Aggregation and Transformation:

We first create a common table expression (CTE) called S. This CTE consolidates medal winners from the Contests table into a single structure, regardless of the medal type (gold, silver, bronze). Each row represents a user's participation in a contest, with user_id representing the user and type denoting the medal type (1 for gold, 2 for silver, 3 for bronze).

2. Consecutive Contest Wins Identification:

The next CTE, T, calculates the difference between the contest_id and the row number for each user, partitioned by user_id and ordered by contest_id. This difference (diff) remains constant for consecutive contests won by the same user.

3. Candidate Identification:

The CTE P identifies the candidates using two conditions:

• Condition 1 (Consecutive Wins): It selects users who have at least 3 consecutive wins (as indicated by a constant diff and a count >= 3 in the T CTE). We leverage GROUP BY and HAVING to filter for users meeting this criteria.
• Condition 2 (Gold Medal Wins): It selects users with at least 3 gold medals from the S CTE using a subquery with COUNT(*) >= 3.

The UNION operator combines the results of these two conditions, ensuring any user satisfying either condition is considered a candidate.

4. Result Retrieval:

Finally, the main query joins P with Users using user_id to retrieve the name and mail of the identified candidates.

Time Complexity Analysis

The time complexity is dominated by the CTEs, specifically T which involves window functions (ROW_NUMBER()). Window functions typically have a time complexity of O(N log N) in the worst case, where N is the number of rows in the S CTE. The GROUP BY and HAVING clauses in P and other CTEs contribute additional logarithmic complexity from sorting, which are negligible compared to the window function. The final JOIN operation has a time complexity that depends on the join algorithm used by the database engine (e.g., hash join or merge join), but it's typically close to O(M log M + N log N) (where M is the number of rows in Users and N in P and assuming appropriate indexing).

Therefore, the overall time complexity is approximately O(N log N), where N is the total number of medal wins recorded across all contests in the Contests table.

Space Complexity Analysis

The space complexity is primarily determined by the size of the CTEs. The size of S is proportional to the number of entries in the Contests table. Similarly, the size of T and P also depends linearly on the number of entries in S. Hence, the space complexity is O(N), where N represents the number of rows in the intermediate CTEs, which is linearly related to the size of the input data.

Code in Other Languages (Conceptual)

While the core logic is expressed in MySQL, the conceptual approach could be implemented in other languages like Python using SQLalchemy or other database interaction libraries. The key steps would remain the same:

• Data Extraction: Retrieve the relevant data from the Contests and Users tables.
• Consecutive Win Detection: Implement logic to identify runs of consecutive wins for each user. This may involve sorting, iteration, and counting.
• Gold Medal Count: Count the gold medals won by each user.
• Candidate Selection: Combine the results from consecutive wins and gold medal counts to identify candidates.
• Output: Display the names and emails of the candidates.

The specific implementation details (e.g., data structures, looping strategies) would vary based on the chosen language and database library. The overall time and space complexities would also be similar to the MySQL solution described above.