How do you solve Find Students Who Improved on LeetCode?

Find Students Who Improved (LeetCode #3421) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using aggregation functions can significantly reduce the complexity of the solution.

What is the time complexity of Find Students Who Improved?

The optimal solution for Find Students Who Improved runs in O(n) time and uses O(n) space. This complexity is linear because we are processing each score once and aggregating results efficiently.

What is the brute force solution for Find Students Who Improved?

The brute force approach for Find Students Who Improved is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every student and their scores in each subject across all exam dates. We will compare the first and latest scores for each subject to determine if there is an improvement. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Find Students Who Improved?

Find Students Who Improved uses the following concepts: Database. The recommended patterns to study are: Hash Map, Aggregation Functions.

What are the interview tips for Find Students Who Improved?

Always think about how to reduce the number of comparisons needed. Make sure to clarify the requirements, such as the need for at least two exam dates. Practice writing SQL queries to become familiar with GROUP BY and HAVING clauses.

What are common mistakes when solving Find Students Who Improved?

Not considering students with only one exam date, leading to incorrect results. Failing to use DISTINCT when counting exam dates, which can lead to inflated counts.

#3421

Find Students Who Improved

Medium

DatabaseHash MapAggregation Functions

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution leverages aggregation functions to efficiently compute the first and latest scores for each student and subject combination in a single pass, reducing unnecessary comparisons.

⚙️

Algorithm

3 steps

1Step 1: Use a GROUP BY clause to group scores by student_id and subject.
2Step 2: Use aggregation functions to find the minimum (first score) and maximum (latest score) for each group.
3Step 3: Filter results using HAVING to ensure at least two distinct exam dates and that the latest score is greater than the first score.

solution.py5 lines

1SELECT student_id, subject, MIN(score) AS first_score, MAX(score) AS latest_score
2FROM Scores
3GROUP BY student_id, subject
4HAVING COUNT(DISTINCT exam_date) >= 2 AND MAX(score) > MIN(score)
5ORDER BY student_id, subject;

ℹ

Complexity note: This complexity is linear because we are processing each score once and aggregating results efficiently.

1Using aggregation functions can significantly reduce the complexity of the solution.
2Filtering results after aggregation helps to ensure we only return relevant data.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.