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

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