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  • LeetCode SQL 50 Solution/1164. Product Price at a Given Date

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Here's the **README.md** file including **MySQL, Window SQL, and Pandas solutions** for **LeetCode 1164 - Product Price at a Given Date**.
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---
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# **1164. Product Price at a Given Date**
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## **Problem Statement**
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You are given the **Products** table, which keeps track of price changes.
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### **Products Table**
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```
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+------------+-----------+-------------+
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| product_id | new_price | change_date |
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+------------+-----------+-------------+
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| int | int | date |
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+------------+-----------+-------------+
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```
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- `(product_id, change_date)` is the **primary key**.
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- Each row represents a price update for a product on a specific date.
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### **Task:**
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Find the price of all products on **2019年08月16日**.
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Assume the **initial price of all products is 10** before any change occurs.
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---
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## **Example 1:**
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### **Input:**
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**Products Table**
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```
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+------------+-----------+-------------+
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| product_id | new_price | change_date |
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+------------+-----------+-------------+
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| 1 | 20 | 2019年08月14日 |
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| 2 | 50 | 2019年08月14日 |
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| 1 | 30 | 2019年08月15日 |
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| 1 | 35 | 2019年08月16日 |
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| 2 | 65 | 2019年08月17日 |
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| 3 | 20 | 2019年08月18日 |
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+------------+-----------+-------------+
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```
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### **Output:**
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```
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+------------+-------+
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| product_id | price |
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+------------+-------+
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| 2 | 50 |
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| 1 | 35 |
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| 3 | 10 |
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+------------+-------+
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```
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### **Explanation:**
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- **Product 1:** Last change before `2019年08月16日` → **35**
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- **Product 2:** Last change before `2019年08月16日` → **50**
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- **Product 3:** **No price change before 2019年08月16日**, so default price is **10**
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---
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## **SQL Solutions**
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### **1️⃣ Standard MySQL Solution**
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```sql
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SELECT
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p.product_id,
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COALESCE((
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SELECT new_price
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FROM Products
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WHERE product_id = p.product_id
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AND change_date <= '2019年08月16日'
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ORDER BY change_date DESC
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LIMIT 1
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), 10) AS price
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FROM
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(SELECT DISTINCT product_id FROM Products) p;
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```
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#### **Explanation:**
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1. **Find the last price before or on `2019年08月16日`**
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- `ORDER BY change_date DESC LIMIT 1` → Gets the most recent price before `2019年08月16日`.
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2. **Use `COALESCE()`**
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- If no price exists, set default price **10**.
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3. **Use `DISTINCT product_id`**
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- Ensures all unique products are checked.
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---
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### **2️⃣ Window Function (SQL) Solution**
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```sql
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WITH RankedPrices AS (
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SELECT
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product_id,
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new_price AS price,
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change_date,
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RANK() OVER (PARTITION BY product_id ORDER BY change_date DESC) AS rnk
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FROM Products
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WHERE change_date <= '2019年08月16日'
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)
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SELECT p.product_id, COALESCE(rp.price, 10) AS price
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FROM (SELECT DISTINCT product_id FROM Products) p
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LEFT JOIN RankedPrices rp ON p.product_id = rp.product_id AND rp.rnk = 1;
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```
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#### **Explanation:**
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1. **`RANK() OVER (PARTITION BY product_id ORDER BY change_date DESC)`**
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- Assigns **rank 1** to the last price before `2019年08月16日`.
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2. **`LEFT JOIN` with `DISTINCT product_id`**
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- Ensures all products are included.
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3. **Use `COALESCE(price, 10)`**
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- If no price exists, set default **10**.
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---
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## **Pandas Solution (Python)**
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```python
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import pandas as pd
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# Sample Data
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products_data = {
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'product_id': [1, 2, 1, 1, 2, 3],
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'new_price': [20, 50, 30, 35, 65, 20],
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'change_date': ['2019年08月14日', '2019年08月14日', '2019年08月15日', '2019年08月16日', '2019年08月17日', '2019年08月18日']
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}
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# Create DataFrame
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products_df = pd.DataFrame(products_data)
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products_df['change_date'] = pd.to_datetime(products_df['change_date']) # Convert to datetime
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# Filter for changes before or on '2019年08月16日'
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valid_prices = products_df[products_df['change_date'] <= '2019年08月16日']
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# Get the latest price for each product before '2019年08月16日'
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latest_prices = valid_prices.sort_values(by=['product_id', 'change_date']).groupby('product_id').last().reset_index()
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# Rename column
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latest_prices = latest_prices[['product_id', 'new_price']].rename(columns={'new_price': 'price'})
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# Get all unique products
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all_products = products_df[['product_id']].drop_duplicates()
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# Merge with latest prices and fill missing values with 10
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final_prices = all_products.merge(latest_prices, on='product_id', how='left').fillna({'price': 10})
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print(final_prices)
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```
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### **Explanation:**
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1. **Convert `change_date` to datetime**
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- Ensures proper date comparison.
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2. **Filter for prices before `2019年08月16日`**
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- Excludes future price changes.
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3. **Get the latest price per product (`groupby().last()`)**
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- Retrieves the most recent price change.
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4. **Merge with all products and set missing prices to `10`**
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- Ensures all products are included.
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---
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## **File Structure**
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```
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LeetCode1164/
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├── problem_statement.md # Contains the problem description and constraints.
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├── sql_solution.sql # Contains the SQL solutions (Standard + Window Functions).
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├── pandas_solution.py # Contains the Pandas solution.
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├── README.md # Overview of the problem and available solutions.
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```
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---
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## **Useful Links**
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- [LeetCode Problem 1164](https://leetcode.com/problems/product-price-at-a-given-date/)
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- [SQL COALESCE Documentation](https://www.w3schools.com/sql/sql_coalesce.asp)
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- [Pandas GroupBy Documentation](https://pandas.pydata.org/docs/reference/api/pandas.DataFrame.groupby.html)
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