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‎LeetCode SQL 50 Solution/1075. Project Employees I/readme.md‎

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Below is a structured README.md for **LeetCode 1174: Immediate Food Delivery II** following your requested format.
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---
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# **1174. Immediate Food Delivery II**
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## **Problem Statement**
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You are given a table `Delivery` that records food deliveries made to customers. Each row represents an order with the date it was placed and the customer’s preferred delivery date.
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---
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## **Delivery Table**
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```
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+-------------+-------------+------------+-----------------------------+
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| Column Name | Type | Description |
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+-------------+-------------+----------------------------------------------+
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| delivery_id | int | Unique identifier for the delivery |
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| customer_id | int | Identifier for the customer |
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| order_date | date | Date when the order was placed |
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| customer_pref_delivery_date | date | Customer’s preferred delivery date |
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+-------------+-------------+----------------------------------------------+
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```
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- `delivery_id` is the **primary key**.
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- Each customer specifies a preferred delivery date, which can be the same as or after the order date.
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---
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## **Task:**
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Calculate the **percentage** of customers whose **first order** is **immediate** (i.e., the order date is the same as the customer’s preferred delivery date).
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- A customer’s **first order** is defined as the order with the **earliest order_date** for that customer.
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- The result should be **rounded to 2 decimal places**.
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- Return the percentage as `immediate_percentage`.
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---
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## **Example 1:**
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### **Input:**
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**Delivery Table**
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```
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+-------------+-------------+------------+-----------------------------+
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| delivery_id | customer_id | order_date | customer_pref_delivery_date |
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+-------------+-------------+------------+-----------------------------+
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| 1 | 1 | 2019-08-01 | 2019-08-02 |
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| 2 | 2 | 2019-08-02 | 2019-08-02 |
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| 3 | 1 | 2019-08-11 | 2019-08-12 |
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| 4 | 3 | 2019-08-24 | 2019-08-24 |
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| 5 | 3 | 2019-08-21 | 2019-08-22 |
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| 6 | 2 | 2019-08-11 | 2019-08-13 |
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| 7 | 4 | 2019-08-09 | 2019-08-09 |
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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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| immediate_percentage |
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+----------------------+
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| 50.00 |
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+----------------------+
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```
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### **Explanation:**
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- **Customer 1:** First order is on **2019-08-01** (preferred: 2019-08-02) → **Scheduled**
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- **Customer 2:** First order is on **2019-08-02** (preferred: 2019-08-02) → **Immediate**
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- **Customer 3:** First order is on **2019-08-21** (preferred: 2019-08-22) → **Scheduled**
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- **Customer 4:** First order is on **2019-08-09** (preferred: 2019-08-09) → **Immediate**
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Out of 4 customers, 2 have immediate first orders.
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Percentage = (2 / 4) * 100 = **50.00**
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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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ROUND(100 * SUM(CASE
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WHEN first_orders.order_date = first_orders.customer_pref_delivery_date THEN 1
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ELSE 0
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END) / COUNT(*), 2) AS immediate_percentage
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FROM (
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-- Get the first order (earliest order_date) for each customer
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SELECT customer_id, order_date, customer_pref_delivery_date
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FROM Delivery
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WHERE (customer_id, order_date) IN (
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SELECT customer_id, MIN(order_date)
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FROM Delivery
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GROUP BY customer_id
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)
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) AS first_orders;
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```
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#### **Explanation:**
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- **Subquery:** Retrieves the first order for each customer by selecting the minimum `order_date`.
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- **Outer Query:**
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- Uses a `CASE` statement to check if the `order_date` equals `customer_pref_delivery_date` (i.e., immediate order).
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- Calculates the percentage of immediate first orders.
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- Rounds the result to 2 decimal places.
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---
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### **2️⃣ Window Function (SQL) Solution**
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```sql
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WITH RankedOrders AS (
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SELECT
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customer_id,
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order_date,
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customer_pref_delivery_date,
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ROW_NUMBER() OVER (PARTITION BY customer_id ORDER BY order_date) AS rn
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FROM Delivery
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)
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SELECT
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ROUND(100 * SUM(CASE WHEN order_date = customer_pref_delivery_date THEN 1 ELSE 0 END) / COUNT(*), 2) AS immediate_percentage
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FROM RankedOrders
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WHERE rn = 1;
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```
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#### **Explanation:**
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- **CTE `RankedOrders`:**
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- Uses `ROW_NUMBER()` to rank orders for each customer by `order_date`.
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- Filters for the first order of each customer (`rn = 1`).
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- **Final SELECT:**
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- Computes the percentage of first orders that are immediate.
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- Rounds the result to 2 decimal places.
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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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def immediate_food_delivery_percentage(delivery: pd.DataFrame) -> pd.DataFrame:
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# Ensure order_date and customer_pref_delivery_date are in datetime format
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delivery['order_date'] = pd.to_datetime(delivery['order_date'])
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delivery['customer_pref_delivery_date'] = pd.to_datetime(delivery['customer_pref_delivery_date'])
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# Get the first order date for each customer
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first_order = delivery.groupby('customer_id')['order_date'].min().reset_index()
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first_order = first_order.rename(columns={'order_date': 'first_order_date'})
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# Merge to get the corresponding preferred delivery date for the first order
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merged = pd.merge(delivery, first_order, on='customer_id', how='inner')
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first_orders = merged[merged['order_date'] == merged['first_order_date']]
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# Calculate immediate orders
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immediate_count = (first_orders['order_date'] == first_orders['customer_pref_delivery_date']).sum()
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total_customers = first_orders['customer_id'].nunique()
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immediate_percentage = round(100 * immediate_count / total_customers, 2)
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return pd.DataFrame({'immediate_percentage': [immediate_percentage]})
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# Example usage:
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# df = pd.read_csv('delivery.csv')
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# print(immediate_food_delivery_percentage(df))
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```
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#### **Explanation:**
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- **Convert Dates:**
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- Convert `order_date` and `customer_pref_delivery_date` to datetime for accurate comparison.
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- **Determine First Order:**
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- Group by `customer_id` to find the minimum `order_date` as the first order.
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- Merge with the original DataFrame to obtain details of the first order.
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- **Calculate Percentage:**
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- Count how many first orders are immediate (where `order_date` equals `customer_pref_delivery_date`).
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- Compute the percentage and round to 2 decimal places.
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---
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## **File Structure**
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```
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LeetCode1174/
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├── problem_statement.md # Contains the problem description and constraints.
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├── sql_standard_solution.sql # Contains the Standard MySQL solution.
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├── sql_window_solution.sql # Contains the Window Function solution.
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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 1174](https://leetcode.com/problems/immediate-food-delivery-ii/)
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- [SQL GROUP BY Documentation](https://www.w3schools.com/sql/sql_groupby.asp)
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- [SQL Window Functions](https://www.w3schools.com/sql/sql_window.asp)
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- [Pandas GroupBy Documentation](https://pandas.pydata.org/docs/reference/api/pandas.DataFrame.groupby.html)
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- [Pandas Merge Documentation](https://pandas.pydata.org/docs/reference/api/pandas.DataFrame.merge.html)
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---
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This complete guide provides multiple solution approaches with detailed explanations, file organization, and useful resources. Happy coding! 🚀

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