test

Author: deepak14ri

ADVANCED SELECT AND JOINS/1731. The Number of Employees Which Report to Each Employee.sql

@@ -1,13 +1,13 @@
-# Write your MySQL query statement below
-SELECT m.employee_id,
-        m.name,
-        count(e.employee_id) AS reports_count,
-        round(avg(e.age)) AS average_age
-FROM
-    employees e JOIN employees m
-ON
-    e.reports_to = m.employee_id
-GROUP BY 
-    employee_id 
-ORDER BY
+# Write your MySQL query statement below
+SELECT m.employee_id,
+        m.name,
+        count(e.employee_id) AS reports_count,
+        round(avg(e.age)) AS average_age
+FROM
+    employees e JOIN employees m
+ON
+    e.reports_to = m.employee_id
+GROUP BY 
+    employee_id 
+ORDER BY
     employee_id;

ADVANCED SELECT AND JOINS/1789. Primary Department for Each Employee.sql

@@ -1,11 +1,11 @@
-# Write your MySQL query statement below
-SELECT employee_id, department_id
-FROM  Employee
-WHERE primary_flag = 'Y'
-    OR
-    employee_id IN (
-        SELECT employee_id
-        FROM Employee
-        GROUP BY employee_id
-        HAVING COUNT(employee_id) = 1
+# Write your MySQL query statement below
+SELECT employee_id, department_id
+FROM  Employee
+WHERE primary_flag = 'Y'
+    OR
+    employee_id IN (
+        SELECT employee_id
+        FROM Employee
+        GROUP BY employee_id
+        HAVING COUNT(employee_id) = 1
     );

ADVANCED SELECT AND JOINS/180. Consecutive Numbers.sql

@@ -1,6 +1,6 @@
--- Write your MySQL/PostgreSQL query statement below
-SELECT DISTINCT l1.num as ConsecutiveNums
-FROM Logs l1
-JOIN Logs l2 ON l1.num = l2.num and l1.id = l2.id + 1
-JOIN Logs l3 ON l2.num = l3.num and l2.id = l3.id + 1;
-
+-- Write your MySQL/PostgreSQL query statement below
+SELECT DISTINCT l1.num as ConsecutiveNums
+FROM Logs l1
+JOIN Logs l2 ON l1.num = l2.num and l1.id = l2.id + 1
+JOIN Logs l3 ON l2.num = l3.num and l2.id = l3.id + 1;
+

ADVANCED SELECT AND JOINS/1907. Count Salary Categories.sql

@@ -1,15 +1,15 @@
-SELECT "Low Salary" AS category,
-       sum(income < 20000) AS accounts_count
-  FROM Accounts
-
-UNION
-
-SELECT "Average Salary" AS category,
-       sum(income BETWEEN 20000 AND 50000) AS accounts_count
-  FROM Accounts
-
-UNION
-
-SELECT "High Salary" AS category,
-       sum(income > 50000) AS accounts_count
+SELECT "Low Salary" AS category,
+       sum(income < 20000) AS accounts_count
+  FROM Accounts
+
+UNION
+
+SELECT "Average Salary" AS category,
+       sum(income BETWEEN 20000 AND 50000) AS accounts_count
+  FROM Accounts
+
+UNION
+
+SELECT "High Salary" AS category,
+       sum(income > 50000) AS accounts_count
   FROM Accounts;

ADVANCED SELECT AND JOINS/610. Triangle Judgement.sql

@@ -1,10 +1,10 @@
-# Write your MySQL query statement below
--- Using the triangle inequality:
-
--- x + y > z
--- x + z > y
--- y + z > x
-
-SELECT x, y, z, CASE WHEN (x+y) > z AND (x+z) > y AND (y+z) > x THEN
-'Yes' ELSE 'No' END AS triangle
-FROM Triangle;
+# Write your MySQL query statement below
+-- Using the triangle inequality:
+
+-- x + y > z
+-- x + z > y
+-- y + z > x
+
+SELECT x, y, z, CASE WHEN (x+y) > z AND (x+z) > y AND (y+z) > x THEN
+'Yes' ELSE 'No' END AS triangle
+FROM Triangle;

ADVANCED STRING FUNCTIONS + REGEX + CLAUSE/1517. Find Users With Valid E-Mails.sql

@@ -1,3 +1,3 @@
-# Write your MySQL query statement below
-SELECT * FROM users
-WHERE mail REGEXP '^[a-zA-Z][a-zA-Z0-9_.-]*@leetcode[.]com$';
+# Write your MySQL query statement below
+SELECT * FROM users
+WHERE mail REGEXP '^[a-zA-Z][a-zA-Z0-9_.-]*@leetcode[.]com$';

ADVANCED STRING FUNCTIONS + REGEX + CLAUSE/1667. Fix Names in a Table.sql

@@ -1,4 +1,4 @@
-# Write your MySQL query statement below
-SELECT user_id, CONCAT(UPPER(LEFT(name, 1)), LOWER(RIGHT(name, LENGTH(name)-1))) AS name
-FROM Users
+# Write your MySQL query statement below
+SELECT user_id, CONCAT(UPPER(LEFT(name, 1)), LOWER(RIGHT(name, LENGTH(name)-1))) AS name
+FROM Users
 ORDER BY user_id;

BASIC AGGREGATION FUNCTION/1075. Project Employees I.sql

@@ -1,5 +1,5 @@
--- Write your PostgreSQL query statement below
-SELECT p.project_id, ROUND(AVG(e.experience_years), 2) AS average_years
-FROM Project p
-LEFT JOIN Employee e ON p.employee_id = e.employee_id
-GROUP BY p.project_id;
+-- Write your PostgreSQL query statement below
+SELECT p.project_id, ROUND(AVG(e.experience_years), 2) AS average_years
+FROM Project p
+LEFT JOIN Employee e ON p.employee_id = e.employee_id
+GROUP BY p.project_id;

BASIC AGGREGATION FUNCTION/1251. Average Selling Price.sql

@@ -1,6 +1,6 @@
-# Write your MySQL query statement below
-SELECT p.product_id, IFNULL(ROUND(SUM(units*price)/SUM(units),2),0) AS average_price
-FROM Prices p LEFT JOIN UnitsSold u
-ON p.product_id = u.product_id AND
-u.purchase_date BETWEEN start_date AND end_date
+# Write your MySQL query statement below
+SELECT p.product_id, IFNULL(ROUND(SUM(units*price)/SUM(units),2),0) AS average_price
+FROM Prices p LEFT JOIN UnitsSold u
+ON p.product_id = u.product_id AND
+u.purchase_date BETWEEN start_date AND end_date
 group by product_id

BASIC AGGREGATION FUNCTION/1633. Percentage of Users Attended a Contest.sql

@@ -1,7 +1,7 @@
-# Write your MySQL query statement below
-SELECT contest_id, ROUND((COUNT(DISTINCT user_id)*100)/(
-    SELECT COUNT(user_id) FROM Users
-), 2) AS percentage
-FROM Register
-GROUP BY contest_id
+# Write your MySQL query statement below
+SELECT contest_id, ROUND((COUNT(DISTINCT user_id)*100)/(
+    SELECT COUNT(user_id) FROM Users
+), 2) AS percentage
+FROM Register
+GROUP BY contest_id
 ORDER BY percentage DESC, contest_id;

BASIC AGGREGATION FUNCTION/620. Not Boring Movies.sql

@@ -1,5 +1,5 @@
-# Write your MySQL query statement below
-SELECT id, movie, description, rating
-FROM Cinema
-WHERE MOD(id, 2) <> 0 AND description <> "boring"
+# Write your MySQL query statement below
+SELECT id, movie, description, rating
+FROM Cinema
+WHERE MOD(id, 2) <> 0 AND description <> "boring"
 ORDER BY rating DESC;

BASIC JOIN/1068. Product Sales Analysis I.sql

@@ -1,3 +1,3 @@
-# Write your MySQL query statement below
-SELECT p.product_name, s.year, s.price
+# Write your MySQL query statement below
+SELECT p.product_name, s.year, s.price
 FROM Sales s LEFT JOIN Product p ON s.product_id = p.product_id;

BASIC JOIN/1378. Replace Employee ID With The Unique Identifier.sql

@@ -1,3 +1,3 @@
-# Write your MySQL query statement below
-SELECT eu.unique_id AS unique_id, e.name AS name
+# Write your MySQL query statement below
+SELECT eu.unique_id AS unique_id, e.name AS name
 FROM Employees e LEFT JOIN EmployeeUNI eu ON e.id = eu.id;

BASIC JOIN/1581. Customer Who Visited but Did Not Make Any Transactions.sql

@@ -1,6 +1,6 @@
-# Write your MySQL query statement below
-SELECT v.customer_id, COUNT(v.visit_id) AS count_no_trans
-FROM Visits v LEFT JOIN Transactions t 
-ON v.visit_id = t.visit_id
-WHERE t.transaction_id IS NULL
+# Write your MySQL query statement below
+SELECT v.customer_id, COUNT(v.visit_id) AS count_no_trans
+FROM Visits v LEFT JOIN Transactions t 
+ON v.visit_id = t.visit_id
+WHERE t.transaction_id IS NULL
 GROUP BY v.customer_id;

BASIC JOIN/197. Rising Temperature.sql

@@ -1,4 +1,4 @@
-# Write your MySQL query statement below
-SELECT W1.id  
-FROM Weather AS W1, Weather AS W2
+# Write your MySQL query statement below
+SELECT W1.id  
+FROM Weather AS W1, Weather AS W2
 WHERE W1.Temperature > W2.Temperature AND DATEDIFF(W1.recordDate, W2.recordDate) = 1;

README.md

@@ -1,44 +1,44 @@
-# LeetCode SQL TOP 50 Solutions
-
-This repository contains solutions to the top 50 SQL problems on LeetCode, covering advanced string functions, regex, clauses, basic aggregation functions, joins, sorting, grouping, subqueries, and more.
-
-## Repository Structure
-
-- `README.md`: An overview of the repository and how to use it.
-- `advanced_string_functions_regex_clauses.sql`: Solutions for advanced string functions, regex, and clause-based problems.
-- `basic_aggregation_function.sql`: Solutions focusing on basic aggregation functions like COUNT, SUM, AVG, etc.
-- `basic_join.sql`: Solutions demonstrating various types of SQL joins (INNER JOIN, LEFT JOIN, RIGHT JOIN, FULL JOIN).
-- `select.sql`: Basic SELECT statement solutions.
-- `sorting_and_group.sql`: Solutions for sorting and grouping data using ORDER BY and GROUP BY.
-- `subqueries.sql`: Solutions involving subqueries.
-
-## How to Use This Repository
-
-1. Clone this repository to your local machine.
-2. Open each SQL file in a text editor or IDE that supports SQL syntax highlighting.
-3. Review the provided solutions and explanations.
-4. Try solving the problems on your own before looking at the solutions.
-5. Use these solutions as a reference or learning tool to improve your SQL skills.
-
-## Latest Commit
-
-**Author:** deepak14ri  
-**Commit Message:** test  
-**Latest Commit Hash:** b6bce78  
-**Date:** 12 hours ago  
-
-## History
-- **SELECT**: Basic SELECT statement solutions to retrieve data from one or more tables.
-- **BASIC AGGREGATION FUNCTION**: Solutions for filtering employee information using basic aggregation functions.
-- **BASIC JOIN**: Demonstrates various join techniques to combine rows from two or more tables based on a related column.
-- **SORTING AND GROUP**: Solutions for sorting and grouping employee information.
-- **SUBQUERIES**: Solutions involving subqueries to perform operations within a query.
-- **ADVANCED STRING FUNCTIONS + REGEX + CLAUSE**: Solutions for problems requiring advanced string manipulation, regular expressions, and specific SQL clauses.
-
-## Contributing
-
-Contributions are welcome Please feel free to submit a pull request if you have improvements or additional solutions.
-
-## License
-
-This project is licensed under the MIT License - see the [LICENSE](LICENSE) file for details.
+# LeetCode SQL TOP 50 Solutions
+
+This repository contains solutions to the top 50 SQL problems on LeetCode, covering advanced string functions, regex, clauses, basic aggregation functions, joins, sorting, grouping, subqueries, and more.
+
+## Repository Structure
+
+- `README.md`: An overview of the repository and how to use it.
+- `advanced_string_functions_regex_clauses.sql`: Solutions for advanced string functions, regex, and clause-based problems.
+- `basic_aggregation_function.sql`: Solutions focusing on basic aggregation functions like COUNT, SUM, AVG, etc.
+- `basic_join.sql`: Solutions demonstrating various types of SQL joins (INNER JOIN, LEFT JOIN, RIGHT JOIN, FULL JOIN).
+- `select.sql`: Basic SELECT statement solutions.
+- `sorting_and_group.sql`: Solutions for sorting and grouping data using ORDER BY and GROUP BY.
+- `subqueries.sql`: Solutions involving subqueries.
+
+## How to Use This Repository
+
+1. Clone this repository to your local machine.
+2. Open each SQL file in a text editor or IDE that supports SQL syntax highlighting.
+3. Review the provided solutions and explanations.
+4. Try solving the problems on your own before looking at the solutions.
+5. Use these solutions as a reference or learning tool to improve your SQL skills.
+
+## Latest Commit
+
+**Author:** deepak14ri  
+**Commit Message:** new problem solved  
+**Latest Commit Hash:** b6bce78  
+**Date:** 12 hours ago  
+
+## History
+- **SELECT**: Basic SELECT statement solutions to retrieve data from one or more tables.
+- **BASIC AGGREGATION FUNCTION**: Solutions for filtering employee information using basic aggregation functions.
+- **BASIC JOIN**: Demonstrates various join techniques to combine rows from two or more tables based on a related column.
+- **SORTING AND GROUP**: Solutions for sorting and grouping employee information.
+- **SUBQUERIES**: Solutions involving subqueries to perform operations within a query.
+- **ADVANCED STRING FUNCTIONS + REGEX + CLAUSE**: Solutions for problems requiring advanced string manipulation, regular expressions, and specific SQL clauses.
+
+## Contributing
+
+Contributions are welcome Please feel free to submit a pull request if you have improvements or additional solutions.
+
+## License
+
+This project is licensed under the MIT License - see the [LICENSE](LICENSE) file for details.

SELECT/1148. Article Views I.sql

@@ -1,4 +1,4 @@
-# Write your MySQL query statement below
-SELECT DISTINCT author_id as id from Views
-WHERE author_id = viewer_id
+# Write your MySQL query statement below
+SELECT DISTINCT author_id as id from Views
+WHERE author_id = viewer_id
 ORDER BY id;

SELECT/1683. Invalid Tweets.sql

@@ -1,2 +1,2 @@
-# Write your MySQL query statement below
+# Write your MySQL query statement below
 SELECT tweet_id FROM Tweets WHERE length(content) > 15;

SELECT/1757. Recyclable and Low Fat Products.sql

@@ -1,5 +1,5 @@
-# Write your MySQL query statement below
-SELECT product_id 
-FROM Products
-WHERE low_fats = 'Y' 
-AND recyclable = 'Y';
+# Write your MySQL query statement below
+SELECT product_id 
+FROM Products
+WHERE low_fats = 'Y' 
+AND recyclable = 'Y';

SELECT/584. Find Customer Referee.sql

@@ -1,3 +1,3 @@
--- Write your PostgreSQL query statement below
-SELECT name from Customer
-WHERE referee_id is null or referee_id != 2;
+-- Write your PostgreSQL query statement below
+SELECT name from Customer
+WHERE referee_id is null or referee_id != 2;

SELECT/595. Big Countries.sql

@@ -1,5 +1,5 @@
-# Write your MySQL query statement below
-SELECT name, population, area
-FROM World
-WHERE area >= 3000000 
+# Write your MySQL query statement below
+SELECT name, population, area
+FROM World
+WHERE area >= 3000000 
 OR population >= 25000000;

SORTING AND GROUP/1070. Product Sales Analysis III.sql

@@ -1,11 +1,11 @@
-# Write your MySQL query statement below
-SELECT product_id, year as first_year, quantity, price
-FROM Sales
-WHERE (product_id, year) 
-IN (SELECT product_id,
-        min(year)
-    FROM Sales
-    GROUP BY
-        product_id
-    );
-
+# Write your MySQL query statement below
+SELECT product_id, year as first_year, quantity, price
+FROM Sales
+WHERE (product_id, year) 
+IN (SELECT product_id,
+        min(year)
+    FROM Sales
+    GROUP BY
+        product_id
+    );
+

SORTING AND GROUP/1141. User Activity for the Past 30 Days I.sql

@@ -1,5 +1,5 @@
-# Write your MySQL query statement below
-SELECT activity_date AS day, COUNT(DISTINCT user_id) AS active_users 
-FROM Activity
-WHERE activity_date BETWEEN '2019-06-28' AND '2019-07-27'
+# Write your MySQL query statement below
+SELECT activity_date AS day, COUNT(DISTINCT user_id) AS active_users 
+FROM Activity
+WHERE activity_date BETWEEN '2019-06-28' AND '2019-07-27'
 GROUP BY activity_date;

SORTING AND GROUP/2356. Number of Unique Subjects Taught by Each Teacher.sql

@@ -1,4 +1,4 @@
-# Write your MySQL query statement below
-SELECT teacher_id, COUNT(DISTINCT subject_id) AS cnt
-FROM Teacher
+# Write your MySQL query statement below
+SELECT teacher_id, COUNT(DISTINCT subject_id) AS cnt
+FROM Teacher
 GROUP BY teacher_id;

SORTING AND GROUP/596. Classes More Than 5 Students.sql

@@ -1,3 +1,3 @@
-# Write your MySQL query statement below
-SELECT class FROM Courses
+# Write your MySQL query statement below
+SELECT class FROM Courses
 GROUP BY class HAVING COUNT(class) >= 5;