Solution Explanation for LeetCode 1623: All Valid Triplets That Can Represent a Country

This problem involves finding all possible triplets of students from three different schools (SchoolA, SchoolB, SchoolC) such that no two students share the same name or ID. The solution uses a SQL query to efficiently achieve this.

Approach

The core idea is to perform a join operation across the three tables (SchoolA, SchoolB, SchoolC) and then filter the results to satisfy the constraints:

1. Join: We use a JOIN (implicitly in this case, since the FROM clause lists all three tables without specifying a JOIN type; it defaults to a CROSS JOIN) to combine all possible combinations of students from the three schools.

2. Filtering: A WHERE clause is used to filter out triplets that violate the constraints:

   • a.student_name != b.student_name: Student names must be distinct.
   • a.student_name != c.student_name: Student names must be distinct.
   • b.student_name != c.student_name: Student names must be distinct.
   • a.student_id != b.student_id: Student IDs must be distinct.
   • a.student_id != c.student_id: Student IDs must be distinct.
   • b.student_id != c.student_id: Student IDs must be distinct.

3. Result Selection: The SELECT statement chooses the student names from each school to form the resulting triplets.

MySQL Code

SELECT
    a.student_name AS member_A,
    b.student_name AS member_B,
    c.student_name AS member_C
FROM
    SchoolA AS a,
    SchoolB AS b,
    SchoolC AS c
WHERE
    a.student_name != b.student_name
    AND a.student_name != c.student_name
    AND b.student_name != c.student_name
    AND a.student_id != b.student_id
    AND a.student_id != c.student_id
    AND b.student_id != c.student_id;

Time Complexity Analysis

The time complexity is dominated by the join operation. In the worst case, a CROSS JOIN will produce a Cartesian product of the number of rows in each table. If n_A, n_B, and n_C are the number of students in SchoolA, SchoolB, and SchoolC respectively, the initial number of combinations is n_A * n_B * n_C. The filtering step then iterates through these combinations, resulting in a time complexity of O(n_A * n_B * n_C). This is because the filtering needs to check each combination, and the number of combinations is the product of the sizes of the tables. Database optimizations might reduce the actual time taken, but the theoretical complexity remains the same.

Space Complexity Analysis

The space complexity is determined by the size of the intermediate result set generated during the join and filtering. In the worst case, this could be as large as n_A * n_B * n_C before filtering, which is then reduced based on how many triplets satisfy the constraints. Thus, the space complexity is also O(n_A * n_B * n_C) in the worst case. However, in practice, the space used will be significantly less due to the filtering conditions. The database management system would optimize the space usage during the query execution.

The solution provided is efficient for this problem because it leverages the database system's capabilities for joining and filtering large datasets. Alternative approaches involving iterating through all combinations in a programming language would be significantly less efficient for larger datasets.