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

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+# 🎓 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)

LeetCode SQL 50 Solution/1321. Restaurant Growth/readme.md

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+# 🍽️ Restaurant Growth - LeetCode 321
+
+## 📌 Problem Statement
+You are given a table **Customer**, which records daily customer transactions in a restaurant. 
+The restaurant owner wants to analyze a **7-day moving average** of customer spending. 
+
+### 📊 Customer Table
+| Column Name | Type |
+| ----------- | ------- |
+| customer_id | int |
+| name | varchar |
+| visited_on | date |
+| amount | int |
+- **(customer_id, visited_on) is the primary key**.
+- `visited_on` represents the date a customer visited the restaurant.
+- `amount` represents the total amount paid by a customer on that day.
+
+---
+
+## 🔢 Goal:
+Compute the **7-day moving average** of customer spending. 
+- The window consists of **current day + 6 days before**.
+- `average_amount` should be **rounded to 2 decimal places**.
+- The result should be **ordered by `visited_on` in ascending order**.
+
+---
+
+## 📊 Example 1:
+### **Input:**
+#### **Customer Table**
+| customer_id | name | visited_on | amount |
+| ----------- | ------- | ---------- | ------ |
+| 1 | Jhon | 2019-01-01 | 100 |
+| 2 | Daniel | 2019-01-02 | 110 |
+| 3 | Jade | 2019-01-03 | 120 |
+| 4 | Khaled | 2019-01-04 | 130 |
+| 5 | Winston | 2019-01-05 | 110 |
+| 6 | Elvis | 2019-01-06 | 140 |
+| 7 | Anna | 2019-01-07 | 150 |
+| 8 | Maria | 2019-01-08 | 80 |
+| 9 | Jaze | 2019-01-09 | 110 |
+| 1 | Jhon | 2019-01-10 | 130 |
+| 3 | Jade | 2019-01-10 | 150 |
+
+### **Output:**
+| visited_on | amount | average_amount |
+| ---------- | ------ | -------------- |
+| 2019-01-07 | 860 | 122.86 |
+| 2019-01-08 | 840 | 120 |
+| 2019-01-09 | 840 | 120 |
+| 2019-01-10 | 1000 | 142.86 |
+
+### **Explanation:**
+1. **First moving average (2019-01-01 to 2019-01-07)** 
+ \[
+ (100 + 110 + 120 + 130 + 110 + 140 + 150) / 7 = 122.86
+ \]
+2. **Second moving average (2019-01-02 to 2019-01-08)** 
+ \[
+ (110 + 120 + 130 + 110 + 140 + 150 + 80) / 7 = 120
+ \]
+3. **Third moving average (2019-01-03 to 2019-01-09)** 
+ \[
+ (120 + 130 + 110 + 140 + 150 + 80 + 110) / 7 = 120
+ \]
+4. **Fourth moving average (2019-01-04 to 2019-01-10)** 
+ \[
+ (130 + 110 + 140 + 150 + 80 + 110 + 130 + 150) / 7 = 142.86
+ \]
+
+---
+
+## 🖥 SQL Solutions
+
+### 1️⃣ **Using `WINDOW FUNCTION` (`SUM() OVER` + `RANK() OVER`)**
+#### **Explanation:**
+- First, **group transactions per day** using `SUM(amount)`.
+- Then, use `SUM() OVER (ROWS 6 PRECEDING)` to calculate **moving sum** over 7 days.
+- Use `RANK()` to track row numbers and filter rows with `rk > 6`.
+- Finally, compute `ROUND(amount / 7, 2)`.
+
+```sql
+WITH t AS (
+ SELECT 
+ visited_on, 
+ SUM(amount) OVER (
+ ORDER BY visited_on 
+ ROWS 6 PRECEDING
+ ) AS amount,
+ RANK() OVER (
+ ORDER BY visited_on 
+ ROWS 6 PRECEDING
+ ) AS rk
+ FROM (
+ SELECT visited_on, SUM(amount) AS amount
+ FROM Customer
+ GROUP BY visited_on
+ ) AS tt
+)
+SELECT visited_on, amount, ROUND(amount / 7, 2) AS average_amount
+FROM t
+WHERE rk > 6;
+```
+
+---
+
+### 2️⃣ **Using `JOIN` + `DATEDIFF()`**
+#### **Explanation:**
+- Use a **self-join** to find transactions **within a 7-day range**.
+- Sum the `amount` for each window and calculate the moving average.
+- Use `DATEDIFF(a.visited_on, b.visited_on) BETWEEN 0 AND 6` to filter records.
+- Ensure only complete 7-day windows are included.
+
+```sql
+SELECT 
+ a.visited_on,
+ SUM(b.amount) AS amount,
+ ROUND(SUM(b.amount) / 7, 2) AS average_amount
+FROM 
+ (SELECT DISTINCT visited_on FROM customer) AS a
+ JOIN customer AS b 
+ ON DATEDIFF(a.visited_on, b.visited_on) BETWEEN 0 AND 6
+WHERE 
+ a.visited_on >= (SELECT MIN(visited_on) FROM customer) + 6
+GROUP BY a.visited_on
+ORDER BY a.visited_on;
+```
+
+---
+
+## 🐍 Pandas Solution (Python)
+#### **Explanation:**
+- **Group by `visited_on`** and sum `amount` per day.
+- **Use `.rolling(7).sum()`** to compute the moving sum over 7 days.
+- **Drop NaN values** to exclude incomplete windows.
+- **Round the average to 2 decimal places**.
+
+```python
+import pandas as pd
+
+def restaurant_growth(customers: pd.DataFrame) -> pd.DataFrame:
+ # Aggregate daily amounts
+ daily_amount = customers.groupby("visited_on")["amount"].sum().reset_index()
+
+ # Compute rolling 7-day sum and moving average
+ daily_amount["amount"] = daily_amount["amount"].rolling(7).sum()
+ daily_amount["average_amount"] = (daily_amount["amount"] / 7).round(2)
+
+ # Drop incomplete windows
+ daily_amount = daily_amount.dropna().reset_index(drop=True)
+
+ return daily_amount
+```
+
+---
+
+## 📁 File Structure
+```
+📂 Restaurant-Growth
+│── 📜 README.md
+│── 📜 solution.sql
+│── 📜 solution_pandas.py
+│── 📜 test_cases.sql
+```
+
+---
+
+## 🔗 Useful Links
+- 📖 [LeetCode Problem](https://leetcode.com/problems/restaurant-growth/)
+- 📚 [SQL `WINDOW FUNCTIONS` Documentation](https://www.w3schools.com/sql/sql_window.asp)
+- 🐍 [Pandas Rolling Window](https://pandas.pydata.org/docs/reference/api/pandas.DataFrame.rolling.html)