Update readme.md

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LeetCode SQL 50 Solution/1280. Students and Examinations/readme.md

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