Solution Explanation

This problem requires finding the optimal cutoff score for each school based on the given capacity and exam results. The solution uses a SQL query to efficiently achieve this.

Approach:

The core idea is to join the Schools and Exam tables based on the condition that the school's capacity is greater than or equal to the number of students achieving a certain score. This ensures that even if all students meeting the cutoff apply, the school can accommodate them. We then select the minimum score that satisfies this condition for each school. If no such score exists, -1 is returned.

MySQL Code:

SELECT school_id, MIN(IFNULL(score, -1)) AS score
FROM
    Schools AS s
    LEFT JOIN Exam AS e ON s.capacity >= e.student_count
GROUP BY school_id;

Code Breakdown:

1. SELECT school_id, MIN(IFNULL(score, -1)) AS score: This selects the school_id and the minimum score (score). IFNULL(score, -1) handles cases where no suitable score is found; in such cases, it returns -1. MIN() ensures that the smallest qualifying score is selected.

2. FROM Schools AS s LEFT JOIN Exam AS e ON s.capacity >= e.student_count: This performs a LEFT JOIN between the Schools table (aliased as s) and the Exam table (aliased as e). The join condition s.capacity >= e.student_count is crucial. It links schools with exam scores where the school's capacity is sufficient to accommodate all students who achieved at least that score. A LEFT JOIN is used to ensure that all schools are included in the result, even if no suitable score is found in the Exam table.

3. GROUP BY school_id: This groups the results by school_id, allowing MIN() to find the minimum score for each school separately.

Time Complexity Analysis:

The time complexity of this SQL query is dominated by the LEFT JOIN operation. In the worst case, the join might need to compare each row in Schools with each row in Exam, resulting in O(M * N) time complexity, where M is the number of rows in Schools and N is the number of rows in Exam. However, database systems typically optimize joins, often achieving better performance in practice. The GROUP BY and MIN() operations have a complexity that is generally less significant compared to the join.

Space Complexity Analysis:

The space complexity is determined by the size of the intermediate result produced by the LEFT JOIN and GROUP BY operations. In the worst case, this could be proportional to the product of the sizes of the input tables (M * N), although efficient database implementations might use optimized data structures to reduce this. The final output size is proportional to the number of schools (M).

Other Languages:

While the problem is naturally suited for SQL, equivalent solutions in other languages would require significantly more code to handle the database interaction, data loading, and processing steps. The core logic of finding the minimum suitable cutoff score would remain similar. The efficiency would largely depend on the chosen data structures and algorithms for processing the data.