|
| 1 | +# Problem 3421: Find Students Who Improved |
| 2 | +# Difficulty: Medium |
| 3 | + |
| 4 | +# Table: Scores |
| 5 | +# +-------------+---------+ |
| 6 | +# | Column Name | Type | |
| 7 | +# +-------------+---------+ |
| 8 | +# | student_id | int | |
| 9 | +# | subject | varchar | |
| 10 | +# | score | int | |
| 11 | +# | exam_date | varchar | |
| 12 | +# +-------------+---------+ |
| 13 | +# (student_id, subject, exam_date) is the primary key. |
| 14 | + |
| 15 | +# Problem Statement: |
| 16 | +# Write a function to find students who have shown improvement. |
| 17 | +# A student is considered to have improved if: |
| 18 | +# - They have taken exams in the same subject on at least two different dates. |
| 19 | +# - Their latest score in that subject is higher than their first score. |
| 20 | +# Return the result table ordered by student_id, subject in ascending order. |
| 21 | + |
| 22 | +# Solution 1 |
| 23 | + |
| 24 | +import pandas as pd |
| 25 | + |
| 26 | +def find_students_who_improved(scores: pd.DataFrame) -> pd.DataFrame: |
| 27 | + # I sort the table by student_id, subject, exam_date so that first and last exams align naturally |
| 28 | + temp_df = scores.sort_values(by=['student_id', 'subject', 'exam_date']) |
| 29 | + |
| 30 | + # I compute first and latest scores by grouping and using 'first' and 'last' |
| 31 | + temp_df1 = temp_df.groupby(['student_id', 'subject'], as_index=False).agg( |
| 32 | + first_score=('score', 'first'), |
| 33 | + latest_score=('score', 'last') |
| 34 | + ) |
| 35 | + |
| 36 | + # I filter out records where the latest score is higher than the first |
| 37 | + result_df = temp_df1[temp_df1['first_score'] < temp_df1['latest_score']] |
| 38 | + |
| 39 | + # I return the final result |
| 40 | + return result_df |
| 41 | + |
| 42 | +# Intuition: |
| 43 | +# I need to compare the first and latest exam scores for each student in each subject and check if the latest score improved. |
| 44 | + |
| 45 | +# Explanation: |
| 46 | +# I first sort the scores by student, subject, and exam date. |
| 47 | +# Then, I group by student and subject to get the first and last scores using 'first' and 'last'. |
| 48 | +# After that, I filter where the latest score is higher than the first and return this result. |
| 49 | + |
| 50 | +# Solution 2 |
| 51 | + |
| 52 | +import pandas as pd |
| 53 | + |
| 54 | +def find_students_who_improved(scores: pd.DataFrame) -> pd.DataFrame: |
| 55 | + # I group by student_id and subject, then compute first and latest scores by date comparison |
| 56 | + temp_df = scores.groupby(['student_id', 'subject'], as_index=False).apply( |
| 57 | + lambda x: pd.Series({ |
| 58 | + 'first_score': x[x['exam_date'] == x['exam_date'].min()]['score'].iloc[0], |
| 59 | + 'latest_score': x[x['exam_date'] == x['exam_date'].max()]['score'].iloc[0] |
| 60 | + }) |
| 61 | + ) |
| 62 | + |
| 63 | + # I filter where the latest score is higher than the first and sort the result |
| 64 | + result_df = temp_df.loc[ |
| 65 | + (temp_df['first_score'] < temp_df['latest_score']) |
| 66 | + ].sort_values(by=['student_id', 'subject'], ascending=[True, True]) |
| 67 | + |
| 68 | + # I return the final result |
| 69 | + return result_df |
| 70 | + |
| 71 | +# Intuition: |
| 72 | +# Similar to Solution 1 — but I explicitly select scores on the earliest and latest exam dates for each student and subject. |
| 73 | + |
| 74 | +# Explanation: |
| 75 | +# I group by student_id and subject, then for each group, find the score on the earliest and latest exam dates. |
| 76 | +# I construct a new DataFrame with these values. |
| 77 | +# Finally, I filter where the latest score is higher than the first, sort by student_id and subject, and return the result. |
0 commit comments