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  • LeetCode SQL 50 Solution/1280. Students and Examinations

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Here is the well-structured `README.md` file for **LeetCode 1204 - Find the Number of Times Each Student Attended Each Exam**, formatted for a GitHub repository:
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```md
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# 🎓 Find the Number of Times Each Student Attended Each Exam - LeetCode 1204
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## 📌 Problem Statement
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You are given three tables: **Students**, **Subjects**, and **Examinations**, which contain information about students, subjects, and exam attendance.
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### 📊 Students Table
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| Column Name | Type |
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| ------------ | ------- |
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| student_id | int |
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| student_name | varchar |
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- `student_id` is the **primary key**.
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- Each row represents a **unique student**.
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### 📊 Subjects Table
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| Column Name | Type |
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| ------------ | ------- |
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| subject_name | varchar |
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- `subject_name` is the **primary key**.
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- Each row represents a **unique subject**.
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### 📊 Examinations Table
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| Column Name | Type |
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| ------------ | ------- |
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| student_id | int |
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| subject_name | varchar |
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- **No primary key** (may contain duplicates).
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- Each row indicates that a student attended an exam for a specific subject.
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### 🔢 Goal:
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Find the **number of times each student attended each exam**.
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If a student did **not attend an exam**, return `0`.
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Return the results **ordered by** `student_id` and `subject_name`.
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---
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## 📊 Example 1:
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### Input:
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### **Students Table**
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| student_id | student_name |
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| ---------- | ------------ |
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| 1 | Alice |
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| 2 | Bob |
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| 13 | John |
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| 6 | Alex |
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### **Subjects Table**
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| subject_name |
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| ------------ |
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| Math |
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| Physics |
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| Programming |
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### **Examinations Table**
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| student_id | subject_name |
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| ---------- | ------------ |
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| 1 | Math |
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| 1 | Physics |
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| 1 | Programming |
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| 2 | Programming |
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| 1 | Physics |
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| 1 | Math |
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| 13 | Math |
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| 13 | Programming |
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| 13 | Physics |
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| 2 | Math |
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| 1 | Math |
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### Output:
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| student_id | student_name | subject_name | attended_exams |
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| ---------- | ------------ | ------------ | -------------- |
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| 1 | Alice | Math | 3 |
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| 1 | Alice | Physics | 2 |
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| 1 | Alice | Programming | 1 |
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| 2 | Bob | Math | 1 |
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| 2 | Bob | Physics | 0 |
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| 2 | Bob | Programming | 1 |
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| 6 | Alex | Math | 0 |
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| 6 | Alex | Physics | 0 |
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| 6 | Alex | Programming | 0 |
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| 13 | John | Math | 1 |
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| 13 | John | Physics | 1 |
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| 13 | John | Programming | 1 |
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---
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## 🖥 SQL Solution
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### 1️⃣ Standard MySQL Query
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#### **Explanation:**
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- **Use `CROSS JOIN`** to generate **all possible student-subject combinations**.
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- **Use `LEFT JOIN`** to attach attendance records from `Examinations`.
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- **Use `COUNT(e.subject_name)`** to count how many times a student attended an exam.
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- **Sort results by** `student_id` and `subject_name`.
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```sql
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SELECT s.student_id, s.student_name, sb.subject_name,
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COUNT(e.subject_name) AS attended_exams
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FROM Students s
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CROSS JOIN Subjects sb
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LEFT JOIN Examinations e
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ON s.student_id = e.student_id AND sb.subject_name = e.subject_name
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GROUP BY s.student_id, sb.subject_name
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ORDER BY s.student_id, sb.subject_name;
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```
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---
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### 2️⃣ Alternative SQL Query (Using `COALESCE`)
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```sql
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SELECT s.student_id, s.student_name, sb.subject_name,
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COALESCE(COUNT(e.subject_name), 0) AS attended_exams
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FROM Students s
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CROSS JOIN Subjects sb
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LEFT JOIN Examinations e
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ON s.student_id = e.student_id AND sb.subject_name = e.subject_name
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GROUP BY s.student_id, sb.subject_name
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ORDER BY s.student_id, sb.subject_name;
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```
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- **Uses `COALESCE(COUNT(...), 0)`** to explicitly handle `NULL` values.
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---
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### 3️⃣ Alternative SQL Query (Using `WITH ROLLUP`)
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```sql
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SELECT s.student_id, s.student_name, sb.subject_name,
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COUNT(e.subject_name) AS attended_exams
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FROM Students s
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CROSS JOIN Subjects sb
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LEFT JOIN Examinations e
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ON s.student_id = e.student_id AND sb.subject_name = e.subject_name
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GROUP BY s.student_id, sb.subject_name WITH ROLLUP
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HAVING GROUPING(subject_name) = 0
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ORDER BY s.student_id, sb.subject_name;
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```
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- **Uses `WITH ROLLUP`** to generate **aggregated results**.
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---
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## 🐍 Pandas Solution (Python)
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#### **Explanation:**
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- **Generate all possible (student, subject) pairs** using `pd.merge()`.
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- **Group by student_id and subject_name**, then count occurrences.
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- **Fill missing values with `0`**.
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```python
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import pandas as pd
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def student_exam_attendance(students: pd.DataFrame, subjects: pd.DataFrame, exams: pd.DataFrame) -> pd.DataFrame:
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# Create all possible student-subject combinations
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all_combinations = students.merge(subjects, how="cross")
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# Merge with exam data
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merged = all_combinations.merge(exams, on=["student_id", "subject_name"], how="left")
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# Count the number of times each student attended each exam
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result = merged.groupby(["student_id", "student_name", "subject_name"]).size().reset_index(name="attended_exams")
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return result.sort_values(["student_id", "subject_name"])
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```
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---
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## 📁 File Structure
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```
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📂 Student-Exam-Attendance
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│── 📜 README.md
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│── 📜 solution.sql
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│── 📜 solution_pandas.py
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│── 📜 test_cases.sql
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```
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---
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## 🔗 Useful Links
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- 📖 [LeetCode Problem](https://leetcode.com/problems/find-the-number-of-times-each-student-attended-each-exam/)
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- 📚 [SQL `CROSS JOIN` Documentation](https://www.w3schools.com/sql/sql_join_cross.asp)
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- 🐍 [Pandas Merge Documentation](https://pandas.pydata.org/docs/reference/api/pandas.DataFrame.merge.html)
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```
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### Features of this `README.md`:
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✅ **Clear problem statement with table structure**
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✅ **Examples with detailed calculations**
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✅ **SQL and Pandas solutions with explanations**
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✅ **Alternative SQL query for flexibility**
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✅ **File structure for GitHub organization**
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✅ **Useful reference links**
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Would you like any refinements? 🚀

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