Table: Schools
+-------------+------+ | Column Name | Type | +-------------+------+ | school_id | int | | capacity | int | +-------------+------+ school_id is the column with unique values for this table. This table contains information about the capacity of some schools. The capacity is the maximum number of students the school can accept.
Table: Exam
+---------------+------+ | Column Name | Type | +---------------+------+ | score | int | | student_count | int | +---------------+------+ score is the column with unique values for this table. Each row in this table indicates that there are student_count students that got at least score points in the exam. The data in this table will be logically correct, meaning a row recording a higher score will have the same or smaller student_count compared to a row recording a lower score. More formally, for every two rows i and j in the table, if scorei > scorej then student_counti <= student_countj.
Every year, each school announces a minimum score requirement that a student needs to apply to it. The school chooses the minimum score requirement based on the exam results of all the students:
Exam table.Write a solution to report the minimum score requirement for each school. If there are multiple score values satisfying the above conditions, choose the smallest one. If the input data is not enough to determine the score, report -1.
Return the result table in any order.
The result format is in the following example.
Example 1:
Input: Schools table: +-----------+----------+ | school_id | capacity | +-----------+----------+ | 11 | 151 | | 5 | 48 | | 9 | 9 | | 10 | 99 | +-----------+----------+ Exam table: +-------+---------------+ | score | student_count | +-------+---------------+ | 975 | 10 | | 966 | 60 | | 844 | 76 | | 749 | 76 | | 744 | 100 | +-------+---------------+ Output: +-----------+-------+ | school_id | score | +-----------+-------+ | 5 | 975 | | 9 | -1 | | 10 | 749 | | 11 | 744 | +-----------+-------+ Explanation: - School 5: The school's capacity is 48. Choosing 975 as the min score requirement, the school will get at most 10 applications, which is within capacity. - School 10: The school's capacity is 99. Choosing 844 or 749 as the min score requirement, the school will get at most 76 applications, which is within capacity. We choose the smallest of them, which is 749. - School 11: The school's capacity is 151. Choosing 744 as the min score requirement, the school will get at most 100 applications, which is within capacity. - School 9: The data given is not enough to determine the min score requirement. Choosing 975 as the min score, the school may get 10 requests while its capacity is 9. We do not have information about higher scores, hence we report -1.
Problem Overview: Each school has a fixed admission capacity. Exam results provide scores and the number of students who achieved each score. For every school, determine the lowest score that still gets admitted when students are taken from highest score downward until the capacity is filled.
Approach 1: Correlated Subquery with Cumulative Counting (O(S * E log E) time, O(1) extra space)
Sort exam scores in descending order and progressively count how many students would be admitted if the cutoff were a particular score. For each school, use a correlated subquery that sums student_count for all rows where score >= current_score. The first score where this cumulative total reaches or exceeds the school's capacity becomes the cutoff. This approach works in pure SQL without advanced features but repeatedly recomputes aggregates, which increases cost as the number of schools or score rows grows.
Approach 2: Window Function with Running Sum (O(E log E + S * E) time, O(E) space)
Compute a running cumulative count of students using a window function. Sort the exam table by score DESC and calculate SUM(student_count) OVER (ORDER BY score DESC). This produces the number of admitted students if the cutoff were each score. Join this derived result with the Schools table and filter rows where the cumulative count is greater than or equal to the school's capacity. Taking the MIN(score) among those candidates gives the exact cutoff. If no cumulative count reaches the capacity, return -1 for that school. Window aggregation avoids repeated scans and is the cleanest solution in modern SQL.
The key insight: admission always proceeds from the highest score downward. Once the cumulative number of students meets or exceeds the capacity, that score defines the cutoff. The window function effectively simulates the admission process in a single pass over sorted scores.
This problem mainly tests reasoning about cumulative aggregates and ranking patterns commonly used in database queries. Window functions like running sums frequently appear in analytics queries and interview-style SQL problems related to window functions.
Recommended for interviews: The window function approach. It demonstrates strong SQL fundamentals, avoids repeated aggregation, and clearly models the admission process with a single ordered pass. The correlated subquery version proves correctness but is less efficient and harder to scale.
MySQL
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Correlated Subquery with Cumulative Count | O(S * E log E) | O(1) | When window functions are unavailable or when working with smaller datasets |
| Window Function Running Sum | O(E log E + S * E) | O(E) | Preferred modern SQL approach for cumulative calculations and analytics queries |
LeetCode Medium 1988 "Find Cutoff Score for Each School" Interview SQL Question with Explanation • Everyday Data Science • 1,912 views views
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