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Merge pull request #90 from iamAntimPal/Branch-1
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  • LeetCode SQL 50 Solution

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

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