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‎LeetCode SQL 50 Solution/584. Find Customer Referee/584. Find Customer Referee.py‎

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import pandas as pd
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def find_customer_referee(customer: pd.DataFrame) -> pd.DataFrame:
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# Filter rows where referee_id is not equal to 2 or is null
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result = customer[(customer['referee_id'] != 2) | (customer['referee_id'].isnull())][['name']]
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return result

‎LeetCode SQL 50 Solution/584. Find Customer Referee/readme.md‎

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# **584. Find Customer Referee**
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## **Problem Statement**
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You are given a table `Customer` that stores customer details along with the ID of the customer who referred them.
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### **Customer Table**
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```
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+-------------+---------+
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| Column Name | Type |
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+-------------+---------+
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| id | int |
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| name | varchar |
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| referee_id | int |
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+-------------+---------+
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```
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- `id` is the **primary key**.
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- Each row represents a customer with their `id`, `name`, and `referee_id`.
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- `referee_id` indicates the customer who referred them. It can be **NULL** if no one referred the customer.
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### **Task:**
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Find the names of the customers who are **not referred** by the customer with `id = 2`.
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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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```
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+----+------+------------+
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| id | name | referee_id |
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+----+------+------------+
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| 1 | Will | null |
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| 2 | Jane | null |
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| 3 | Alex | 2 |
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| 4 | Bill | null |
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| 5 | Zack | 1 |
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| 6 | Mark | 2 |
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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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| name |
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+------+
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| Will |
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| Jane |
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| Bill |
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| Zack |
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+------+
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```
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### **Explanation:**
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- **Alex** (id = 3) and **Mark** (id = 6) are referred by the customer with `id = 2` and are excluded.
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- The remaining customers (**Will**, **Jane**, **Bill**, **Zack**) are not referred by the customer with `id = 2`.
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---
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## **Solution Approaches**
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### **SQL Solution (Using WHERE Clause)**
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```sql
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SELECT name
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FROM Customer
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WHERE referee_id != 2 OR referee_id IS NULL;
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```
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**Explanation:**
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- The query selects customer names where `referee_id` is either not equal to `2` or is `NULL`.
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- This effectively filters out customers referred by the customer with `id = 2`.
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---
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### **Pandas Solution**
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```python
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import pandas as pd
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def find_customer_referee(customer: pd.DataFrame) -> pd.DataFrame:
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# Filter rows where referee_id is not equal to 2 or is null
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result = customer[(customer['referee_id'] != 2) | (customer['referee_id'].isnull())][['name']]
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return result
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```
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**Explanation:**
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- The Pandas solution filters the DataFrame for rows where `referee_id` is not 2 or is `NaN`.
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- It then returns the `name` column containing the desired customer names.
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---
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## **File Structure**
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```
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LeetCode584/
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├── problem_statement.md # Contains the problem description and constraints.
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├── sql_solution.sql # Contains the SQL solution.
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├── pandas_solution.py # Contains the Pandas solution.
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├── README.md # Overview of the problem and solutions.
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```
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---
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## **Useful Links**
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- [LeetCode Problem 584](https://leetcode.com/problems/find-customer-referee/)
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- [SQL WHERE Clause Documentation](https://www.w3schools.com/sql/sql_where.asp)
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- [Pandas Filtering DataFrames](https://pandas.pydata.org/docs/reference/api/pandas.DataFrame.html)
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‎LeetCode SQL 50 Solution/585. Investments in 2016/585. Investments in 2016.py‎

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import pandas as pd
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def investments_in_2016(insurance: pd.DataFrame) -> pd.DataFrame:
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# Count the number of occurrences for each tiv_2015 value
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insurance['tiv_2015_count'] = insurance.groupby('tiv_2015')['tiv_2015'].transform('count')
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# Count the number of occurrences for each (lat, lon) pair
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insurance['city_count'] = insurance.groupby(['lat', 'lon'])['lat'].transform('count')
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# Filter rows that meet both criteria:
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# 1. tiv_2015 appears more than once.
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# 2. The location (lat, lon) is unique (appears only once).
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valid_rows = insurance[(insurance['tiv_2015_count'] > 1) & (insurance['city_count'] == 1)]
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# Calculate the sum of tiv_2016 and round to 2 decimal places
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total_tiv_2016 = round(valid_rows['tiv_2016'].sum(), 2)
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# Return result as a DataFrame
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return pd.DataFrame({'tiv_2016': [total_tiv_2016]})

‎LeetCode SQL 50 Solution/585. Investments in 2016/readme.md‎

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# **585. Investments in 2016**
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## **Problem Statement**
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You are given a table `Insurance` that contains details about policy investments.
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### **Insurance Table**
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```
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+-------------+-------+
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| Column Name | Type |
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+-------------+-------+
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| pid | int |
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| tiv_2015 | float |
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| tiv_2016 | float |
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| lat | float |
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| lon | float |
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+-------------+-------+
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```
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- `pid` is the **primary key**.
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- `tiv_2015` is the total investment value in **2015**.
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- `tiv_2016` is the total investment value in **2016**.
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- `lat` and `lon` represent the **latitude** and **longitude** of the policyholder's city. Both values are guaranteed not to be `NULL`.
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### **Task:**
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Report the **sum** of all `tiv_2016` values (rounded to two decimal places) for all policyholders who:
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1. Have the **same `tiv_2015`** value as one or more other policyholders.
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2. Are **not located** in the same city as any other policyholder (i.e., the `(lat, lon)` attribute pair is **unique**).
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---
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## **Example 1:**
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### **Input:**
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#### **Insurance Table**
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```
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+-----+----------+----------+-----+-----+
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| pid | tiv_2015 | tiv_2016 | lat | lon |
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+-----+----------+----------+-----+-----+
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| 1 | 10 | 5 | 10 | 10 |
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| 2 | 20 | 20 | 20 | 20 |
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| 3 | 10 | 30 | 20 | 20 |
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| 4 | 10 | 40 | 40 | 40 |
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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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| tiv_2016 |
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+----------+
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| 45.00 |
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+----------+
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```
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### **Explanation:**
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- The policyholders with `tiv_2015 = 10` appear in multiple rows.
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- Among these, only the records with **unique locations** count.
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- Policy `pid = 1` and `pid = 4` meet both criteria, so the result is the sum of their `tiv_2016`: **5 + 40 = 45.00**.
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---
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## **Solution Approaches**
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### **SQL Solution (Using Window Functions)**
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```sql
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WITH InsuranceWithCounts AS (
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SELECT
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tiv_2016,
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COUNT(*) OVER(PARTITION BY tiv_2015) AS tiv_2015_count,
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COUNT(*) OVER(PARTITION BY lat, lon) AS city_count
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FROM Insurance
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)
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SELECT ROUND(SUM(tiv_2016), 2) AS tiv_2016
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FROM InsuranceWithCounts
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WHERE tiv_2015_count > 1
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AND city_count = 1;
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```
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**Explanation:**
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- The CTE `InsuranceWithCounts` computes:
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- `tiv_2015_count`: Number of records with the same `tiv_2015`.
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- `city_count`: Number of records with the same `(lat, lon)` pair.
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- The outer query filters rows where:
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- `tiv_2015_count > 1` (i.e., policyholders share their 2015 investment value with others).
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- `city_count = 1` (i.e., their location is unique).
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- Finally, it sums `tiv_2016` and rounds the result to 2 decimal places.
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---
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### **Pandas Solution**
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```python
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import pandas as pd
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def investments_in_2016(insurance: pd.DataFrame) -> pd.DataFrame:
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# Count the number of occurrences for each tiv_2015 value
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insurance['tiv_2015_count'] = insurance.groupby('tiv_2015')['tiv_2015'].transform('count')
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# Count the number of occurrences for each (lat, lon) pair
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insurance['city_count'] = insurance.groupby(['lat', 'lon'])['lat'].transform('count')
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# Filter rows that meet both criteria:
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# 1. tiv_2015 appears more than once.
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# 2. The location (lat, lon) is unique (appears only once).
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valid_rows = insurance[(insurance['tiv_2015_count'] > 1) & (insurance['city_count'] == 1)]
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# Calculate the sum of tiv_2016 and round to 2 decimal places
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total_tiv_2016 = round(valid_rows['tiv_2016'].sum(), 2)
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# Return result as a DataFrame
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return pd.DataFrame({'tiv_2016': [total_tiv_2016]})
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# Example usage:
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# df = pd.read_csv('insurance.csv')
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# print(investments_in_2016(df))
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```
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**Explanation:**
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- The code computes two new columns:
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- `tiv_2015_count` for the number of policyholders with the same 2015 investment.
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- `city_count` for the count of policyholders in each unique city (using `(lat, lon)`).
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- Rows that meet the conditions are filtered.
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- The `tiv_2016` values of the valid rows are summed and rounded to 2 decimal places.
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- The result is returned as a DataFrame.
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---
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## **File Structure**
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```
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LeetCode585/
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├── problem_statement.md # Contains the problem description and constraints.
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├── sql_solution.sql # SQL solution using window functions.
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├── pandas_solution.py # Pandas solution for Python users.
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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 585](https://leetcode.com/problems/investments-in-2016/)
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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 Transform Documentation](https://pandas.pydata.org/docs/reference/api/pandas.DataFrame.transform.html)

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