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LeetCode SQL Problem Solving Questions With Solutions

LeetCode SQL Problem Solving Questions With Solutions

175. Combine Two Tables | Easy | LeetCode

Table: Person

+-------------+---------+
| Column Name | Type |
+-------------+---------+
| PersonId | int |
| FirstName | varchar |
| LastName | varchar |
+-------------+---------+
PersonId is the primary key column for this table.

Table: Address

+-------------+---------+
| Column Name | Type |
+-------------+---------+
| AddressId | int |
| PersonId | int |
| City | varchar |
| State | varchar |
+-------------+---------+
AddressId is the primary key column for this table.

Write a SQL query for a report that provides the following information for each person in the Person table, regardless if there is an address for each of those people:

FirstName, LastName, City, State

Solution

sql
SELECT p.FirstName, p.LastName, a.City, a.State
FROM Person p
LEFT JOIN Address a
ON p.PersonId = a.PersonId;

176. Second Highest Salary | Easy | LeetCode

Write a SQL query to get the second highest salary from the Employee table.

+----+--------+
| Id | Salary |
+----+--------+
| 1 | 100 |
| 2 | 200 |
| 3 | 300 |
+----+--------+

For example, given the above Employee table, the query should return 200 as the second highest salary. If there is no second highest salary, then the query should return null.

+---------------------+
| SecondHighestSalary |
+---------------------+
| 200 |
+---------------------+

Solution

sql
#Solution 1:
SELECT Max(Salary) SecondHighestSalary
FROM Employee WHERE Salary < (SELECT MAX(Salary) FROM Employee)
#Solution 2:
WITH CTE AS (SELECT DISTINCT Salary
FROM Employee
ORDER BY Salary DESC
LIMIT 2)
SELECT Salary as SecondHighestSalary
FROM CTE
ORDER BY Salary Asc
LIMIT 1;
#Solution 3:
WITH CTE AS
(
SELECT Salary,
DENSE_RANK() OVER (ORDER BY Salary DESC) AS DENSERANK
FROM Employee
)
SELECT Salary SecondHighestSalary
FROM CTE
WHERE DENSERANK = 2;

177. Nth Highest Salary | Medium | LeetCode

Write a SQL query to get the nth highest salary from the Employee table.

+----+--------+
| Id | Salary |
+----+--------+
| 1 | 100 |
| 2 | 200 |
| 3 | 300 |
+----+--------+

For example, given the above Employee table, the nth highest salary where n = 2 is 200. If there is no nth highest salary, then the query should return null.

+------------------------+
| getNthHighestSalary(2) |
+------------------------+
| 200 |
+------------------------+

Solution

sql
CREATE FUNCTION getNthHighestSalary(N INT) RETURNS INT
BEGIN
SET N = N-1;
RETURN(
SELECT DISTINCT Salary FROM Employee ORDER BY Salary DESC
LIMIT 1 OFFSET N
);
END

178. Rank Scores | Medium | LeetCode

Write a SQL query to rank scores. If there is a tie between two scores, both should have the same ranking. Note that after a tie, the next ranking number should be the next consecutive integer value. In other words, there should be no "holes" between ranks.

+----+-------+
| Id | Score |
+----+-------+
| 1 | 3.50 |
| 2 | 3.65 |
| 3 | 4.00 |
| 4 | 3.85 |
| 5 | 4.00 |
| 6 | 3.65 |
+----+-------+

For example, given the above Scores table, your query should generate the following report (order by highest score):

+-------+---------+
| score | Rank |
+-------+---------+
| 4.00 | 1 |
| 4.00 | 1 |
| 3.85 | 2 |
| 3.65 | 3 |
| 3.65 | 3 |
| 3.50 | 4 |
+-------+---------+

Important Note: For MySQL solutions, to escape reserved words used as column names, you can use an apostrophe before and after the keyword. For example Rank.

Solution

sql
SELECT score, DENSE_RANK() OVER (ORDER By Score DESC) AS "Rank"
FROM Scores;

180. Consecutive Numbers | Medium | LeetCode

Table: Logs

+-------------+---------+
| Column Name | Type |
+-------------+---------+
| id | int |
| num | varchar |
+-------------+---------+
id is the primary key for this table.

Write an SQL query to find all numbers that appear at least three times consecutively.

Return the result table in any order.

The query result format is in the following example:

Logs table:
+----+-----+
| Id | Num |
+----+-----+
| 1 | 1 |
| 2 | 1 |
| 3 | 1 |
| 4 | 2 |
| 5 | 1 |
| 6 | 2 |
| 7 | 2 |
+----+-----+
Result table:
+-----------------+
| ConsecutiveNums |
+-----------------+
| 1 |
+-----------------+
1 is the only number that appears consecutively for at least three times.

Solution

sql
SELECT a.Num as ConsecutiveNums
FROM Logs a
JOIN Logs b
ON a.id = b.id+1 AND a.num = b.num
JOIN Logs c
ON a.id = c.id+2 AND a.num = c.num;

181. Employees Earning More Than Their Managers | Easy | LeetCode

The Employee table holds all employees including their managers. Every employee has an Id, and there is also a column for the manager Id.

+----+-------+--------+-----------+
| Id | Name | Salary | ManagerId |
+----+-------+--------+-----------+
| 1 | Joe | 70000 | 3 |
| 2 | Henry | 80000 | 4 |
| 3 | Sam | 60000 | NULL |
| 4 | Max | 90000 | NULL |
+----+-------+--------+-----------+

Given the Employee table, write a SQL query that finds out employees who earn more than their managers. For the above table, Joe is the only employee who earns more than his manager.

+----------+
| Employee |
+----------+
| Joe |
+----------+

Solution

sql
SELECT E.Name as "Employee"
FROM Employee E
JOIN Employee M
ON E.ManagerId = M.Id
AND E.Salary > M.Salary;

182. Duplicate Emails | Easy | LeetCode

Write a SQL query to find all duplicate emails in a table named Person.

+----+---------+
| Id | Email |
+----+---------+
| 1 | a@b.com |
| 2 | c@d.com |
| 3 | a@b.com |
+----+---------+

For example, your query should return the following for the above table:

+---------+
| Email |
+---------+
| a@b.com |
+---------+

Note: All emails are in lowercase.

Solution

sql
#Solution- 1:
SELECT Email
FROM Person
GROUP BY Email
HAVING count(*) > 1
#Solution- 2:
WITH CTE AS(
SELECT Email, ROW_NUMBER() OVER(PARTITION BY Email ORDER BY Email) AS RN
FROM Person
)
SELECT Email
FROM CTE
WHERE RN > 1;

183. Customers Who Never Order | Easy | LeetCode

Suppose that a website contains two tables, the Customers table and the Orders table. Write a SQL query to find all customers who never order anything.

Table: Customers.

+----+-------+
| Id | Name |
+----+-------+
| 1 | Joe |
| 2 | Henry |
| 3 | Sam |
| 4 | Max |
+----+-------+

Table: Orders.

+----+------------+
| Id | CustomerId |
+----+------------+
| 1 | 3 |
| 2 | 1 |
+----+------------+

Using the above tables as example, return the following:

+-----------+
| Customers |
+-----------+
| Henry |
| Max |
+-----------+

Solution

sql
#Solution- 1:
SELECT Name AS Customers
FROM Customers
LEFT JOIN Orders
ON Customers.Id = Orders.CustomerId
WHERE CustomerId IS NULL;
#Solution- 2:
SELECT Name as Customers
FROM Customers
WHERE Id NOT IN(
SELECT CustomerId
FROM Orders
)

184. Department Highest Salary | Medium | LeetCode

The Employee table holds all employees. Every employee has an Id, a salary, and there is also a column for the department Id.

+----+-------+--------+--------------+
| Id | Name | Salary | DepartmentId |
+----+-------+--------+--------------+
| 1 | Joe | 70000 | 1 |
| 2 | Jim | 90000 | 1 |
| 3 | Henry | 80000 | 2 |
| 4 | Sam | 60000 | 2 |
| 5 | Max | 90000 | 1 |
+----+-------+--------+--------------+

The Department table holds all departments of the company.

+----+----------+
| Id | Name |
+----+----------+
| 1 | IT |
| 2 | Sales |
+----+----------+

Write a SQL query to find employees who have the highest salary in each of the departments. For the above tables, your SQL query should return the following rows (order of rows does not matter).

+------------+----------+--------+
| Department | Employee | Salary |
+------------+----------+--------+
| IT | Max | 90000 |
| IT | Jim | 90000 |
| Sales | Henry | 80000 |
+------------+----------+--------+

Explanation:
Max and Jim both have the highest salary in the IT department and Henry has the highest salary in the Sales department.

Solution

sql
SELECT Department.Name AS Department, Employee.Name AS Employee, Salary
FROM Employee
JOIN Department
ON Employee.DepartmentId = Department.Id
WHERE (DepartmentId, Salary) IN(
SELECT DepartmentId, MAX(Salary) AS Salary
FROM Employee
GROUP BY DepartmentId
);

185. Department Top Three Salaries | Hard | LeetCode

The Employee table holds all employees. Every employee has an Id, and there is also a column for the department Id.

+----+-------+--------+--------------+
| Id | Name` | Salary | DepartmentId |
+----+-------+--------+--------------+
| 1 | Joe | 85000 | 1 |
| 2 | Henry | 80000 | 2 |
| 3 | Sam | 60000 | 2 |
| 4 | Max | 90000 | 1 |
| 5 | Janet | 69000 | 1 |
| 6 | Randy | 85000 | 1 |
| 7 | Will | 70000 | 1 |
+----+-------+--------+--------------+

The Department table holds all departments of the company.

+----+----------+
| Id | Name |
+----+----------+
| 1 | IT |
| 2 | Sales |
+----+----------+

Write a SQL query to find employees who earn the top three salaries in each of the department. For the above tables, your SQL query should return the following rows (order of rows does not matter).

+------------+----------+--------+
| Department | Employee | Salary |
+------------+----------+--------+
| IT | Max | 90000 |
| IT | Randy | 85000 |
| IT | Joe | 85000 |
| IT | Will | 70000 |
| Sales | Henry | 80000 |
| Sales | Sam | 60000 |
+------------+----------+--------+

Explanation:
In IT department, Max earns the highest salary, both Randy and Joe earn the second highest salary, and Will earns the third highest salary. There are only two employees in the Sales department, Henry earns the highest salary while Sam earns the second highest salary.

Solution

sql
WITH department_ranking AS (
SELECT Name AS Employee, Salary ,DepartmentId
,DENSE_RANK() OVER (PARTITION BY DepartmentId ORDER BY Salary DESC) AS rnk
FROM Employee
)
SELECT d.Name AS Department, r.Employee, r.Salary
FROM department_ranking AS r
JOIN Department AS d
ON r.DepartmentId = d.Id
WHERE r.rnk <= 3
ORDER BY d.Name ASC, r.Salary DESC;

196. Delete Duplicate Emails | Easy | LeetCode

Write a SQL query to delete all duplicate email entries in a table named Person, keeping only unique emails based on its smallest Id.

+----+------------------+
| Id | Email |
+----+------------------+
| 1 | john@example.com |
| 2 | bob@example.com |
| 3 | john@example.com |
+----+------------------+

Id is the primary key column for this table. For example, after running your query, the above Person table should have the following rows:

+----+------------------+
| Id | Email |
+----+------------------+
| 1 | john@example.com |
| 2 | bob@example.com |
+----+------------------+

Note:
Your output is the whole Person table after executing your sql. Use delete statement.

Solution

sql
DELETE p2
FROM Person p1
JOIN Person p2
ON p1.Email = p2.Email
AND p1.id < p2.id

197. Rising Temperature | Easy | LeetCode

Table: Weather

+---------------+---------+
| Column Name | Type |
+---------------+---------+
| id | int |
| recordDate | date |
| temperature | int |
+---------------+---------+
id is the primary key for this table.
This table contains information about the temperature in a certain day.

Write an SQL query to find all dates' id with higher temperature compared to its previous dates (yesterday).

Return the result table in any order.

The query result format is in the following example:

Weather
+----+------------+-------------+
| id | recordDate | Temperature |
+----+------------+-------------+
| 1 | 2015-01-01 | 10 |
| 2 | 2015-01-02 | 25 |
| 3 | 2015-01-03 | 20 |
| 4 | 2015-01-04 | 30 |
+----+------------+-------------+
Result table:
+----+
| id |
+----+
| 2 |
| 4 |
+----+
In 2015-01-02, temperature was higher than the previous day (10 -> 25).
In 2015-01-04, temperature was higher than the previous day (20 -> 30).

Solution

sql
#Solution- 1:
SELECT t.Id
FROM Weather AS t, Weather AS y
WHERE DATEDIFF(t.RecordDate, y.RecordDate) = 1
AND t.Temperature > y.Temperature;
#Solution- 2:
SELECT t.Id
FROM Weather t
JOIN Weather y
ON DATEDIFF(t.recordDate, y.recordDate) = 1 AND
t.temperature > y.temperature;

262. Trips and Users | Hard | LeetCode

Table: Trips

+-------------+----------+
| Column Name | Type |
+-------------+----------+
| Id | int |
| Client_Id | int |
| Driver_Id | int |
| City_Id | int |
| Status | enum |
| Request_at | date |
+-------------+----------+
Id is the primary key for this table.
The table holds all taxi trips. Each trip has a unique Id, while Client_Id and Driver_Id are foreign keys to the Users_Id at the Users table.
Status is an ENUM type of (β€˜completed’, β€˜cancelled_by_driver’, β€˜cancelled_by_client’).

Table: Users

+-------------+----------+
| Column Name | Type |
+-------------+----------+
| Users_Id | int |
| Banned | enum |
| Role | enum |
+-------------+----------+
Users_Id is the primary key for this table.
The table holds all users. Each user has a unique Users_Id, and Role is an ENUM type of (β€˜client’, β€˜driver’, β€˜partner’).
Status is an ENUM type of (β€˜Yes’, β€˜No’).

Write a SQL query to find the cancellation rate of requests with unbanned users (both client and driver must not be banned) each day between "2013-10-01" and "2013-10-03".

The cancellation rate is computed by dividing the number of canceled (by client or driver) requests with unbanned users by the total number of requests with unbanned users on that day.

Return the result table in any order. Round Cancellation Rate to two decimal points.

The query result format is in the following example:

Trips table:
+----+-----------+-----------+---------+---------------------+------------+
| Id | Client_Id | Driver_Id | City_Id | Status | Request_at |
+----+-----------+-----------+---------+---------------------+------------+
| 1 | 1 | 10 | 1 | completed | 2013-10-01 |
| 2 | 2 | 11 | 1 | cancelled_by_driver | 2013-10-01 |
| 3 | 3 | 12 | 6 | completed | 2013-10-01 |
| 4 | 4 | 13 | 6 | cancelled_by_client | 2013-10-01 |
| 5 | 1 | 10 | 1 | completed | 2013-10-02 |
| 6 | 2 | 11 | 6 | completed | 2013-10-02 |
| 7 | 3 | 12 | 6 | completed | 2013-10-02 |
| 8 | 2 | 12 | 12 | completed | 2013-10-03 |
| 9 | 3 | 10 | 12 | completed | 2013-10-03 |
| 10 | 4 | 13 | 12 | cancelled_by_driver | 2013-10-03 |
+----+-----------+-----------+---------+---------------------+------------+
Users table:
+----------+--------+--------+
| Users_Id | Banned | Role |
+----------+--------+--------+
| 1 | No | client |
| 2 | Yes | client |
| 3 | No | client |
| 4 | No | client |
| 10 | No | driver |
| 11 | No | driver |
| 12 | No | driver |
| 13 | No | driver |
+----------+--------+--------+
Result table:
+------------+-------------------+
| Day | Cancellation Rate |
+------------+-------------------+
| 2013-10-01 | 0.33 |
| 2013-10-02 | 0.00 |
| 2013-10-03 | 0.50 |
+------------+-------------------+
On 2013-10-01:
- There were 4 requests in total, 2 of which were canceled.
- However, the request with Id=2 was made by a banned client (User_Id=2), so it is ignored in the calculation.
- Hence there are 3 unbanned requests in total, 1 of which was canceled.
- The Cancellation Rate is (1 / 3) = 0.33
On 2013-10-02:
- There were 3 requests in total, 0 of which were canceled.
- The request with Id=6 was made by a banned client, so it is ignored.
- Hence there are 2 unbanned requests in total, 0 of which were canceled.
- The Cancellation Rate is (0 / 2) = 0.00
On 2013-10-03:
- There were 3 requests in total, 1 of which was canceled.
- The request with Id=8 was made by a banned client, so it is ignored.
- Hence there are 2 unbanned request in total, 1 of which were canceled.
- The Cancellation Rate is (1 / 2) = 0.50

Solution

sql
SELECT Request_at AS Day,
ROUND(SUM(IF(Status<>"completed", 1, 0))/COUNT(Status),2) AS "Cancellation Rate"
FROM Trips
WHERE Request_at BETWEEN "2013-10-01" AND "2013-10-03"
AND Client_Id NOT IN (SELECT Users_Id FROM Users WHERE Banned = 'Yes')
AND Driver_Id NOT IN (SELECT Users_Id FROM Users WHERE Banned = 'Yes')
GROUP BY Request_at;

511. Game Play Analysis I | Easy | πŸ”’ LeetCode

Table: Activity

+--------------+---------+
| Column Name | Type |
+--------------+---------+
| player_id | int |
| device_id | int |
| event_date | date |
| games_played | int |
+--------------+---------+
(player_id, event_date) is the primary key of this table.
This table shows the activity of players of some game.
Each row is a record of a player who logged in and played a number of games (possibly 0) before logging out on some day using some device.

Write an SQL query that reports the first login date for each player.

The query result format is in the following example:

Activity table:
+-----------+-----------+------------+--------------+
| player_id | device_id | event_date | games_played |
+-----------+-----------+------------+--------------+
| 1 | 2 | 2016-03-01 | 5 |
| 1 | 2 | 2016-05-02 | 6 |
| 2 | 3 | 2017-06-25 | 1 |
| 3 | 1 | 2016-03-02 | 0 |
| 3 | 4 | 2018-07-03 | 5 |
+-----------+-----------+------------+--------------+
Result table:
+-----------+-------------+
| player_id | first_login |
+-----------+-------------+
| 1 | 2016-03-01 |
| 2 | 2017-06-25 |
| 3 | 2016-03-02 |
+-----------+-------------+

Solution

sql
SELECT player_id, MIN(event_date) as first_login
FROM Activity
GROUP BY player_id

512. Game Play Analysis II | Easy | πŸ”’ LeetCode

Table: Activity

+--------------+---------+
| Column Name | Type |
+--------------+---------+
| player_id | int |
| device_id | int |
| event_date | date |
| games_played | int |
+--------------+---------+
(player_id, event_date) is the primary key of this table.
This table shows the activity of players of some game.
Each row is a record of a player who logged in and played a number of games (possibly 0) before logging out on some day using some device.

Write a SQL query that reports the device that is first logged in for each player.

The query result format is in the following example:

Activity table:
+-----------+-----------+------------+--------------+
| player_id | device_id | event_date | games_played |
+-----------+-----------+------------+--------------+
| 1 | 2 | 2016-03-01 | 5 |
| 1 | 2 | 2016-05-02 | 6 |
| 2 | 3 | 2017-06-25 | 1 |
| 3 | 1 | 2016-03-02 | 0 |
| 3 | 4 | 2018-07-03 | 5 |
+-----------+-----------+------------+--------------+
Result table:
+-----------+-----------+
| player_id | device_id |
+-----------+-----------+
| 1 | 2 |
| 2 | 3 |
| 3 | 1 |
+-----------+-----------+

Solution

sql
#Solution- 1:
SELECT DISTINCT player_id, device_id
FROM Activity
WHERE (player_id, event_date) in (
SELECT player_id, min(event_date)
FROM Activity
GROUP BY player_id)
#Solution- 2:
SELECT a.player_id, b.device_id
FROM
(SELECT player_id, MIN(event_date) AS event_date FROM Activity
GROUP BY player_id) a
JOIN Activity b
ON a.player_id = b.player_id AND a.event_date = b.event_date;
#Solution- 3:
SELECT player_id, device_id
FROM
(SELECT player_id, device_id, event_date,
ROW_NUMBER() OVER (PARTITION BY player_id ORDER BY event_date) AS r
FROM Activity) lookup
WHERE r = 1;

534. Game Play Analysis III | Medium | πŸ”’ LeetCode

Table: Activity

+--------------+---------+
| Column Name | Type |
+--------------+---------+
| player_id | int |
| device_id | int |
| event_date | date |
| games_played | int |
+--------------+---------+
(player_id, event_date) is the primary key of this table.
This table shows the activity of players of some game.
Each row is a record of a player who logged in and played a number of games (possibly 0) before logging out on some day using some device.

Write an SQL query that reports for each player and date, how many games played so far by the player. That is, the total number of games played by the player until that date. Check the example for clarity.

The query result format is in the following example:

Activity table:
+-----------+-----------+------------+--------------+
| player_id | device_id | event_date | games_played |
+-----------+-----------+------------+--------------+
| 1 | 2 | 2016-03-01 | 5 |
| 1 | 2 | 2016-05-02 | 6 |
| 1 | 3 | 2017-06-25 | 1 |
| 3 | 1 | 2016-03-02 | 0 |
| 3 | 4 | 2018-07-03 | 5 |
+-----------+-----------+------------+--------------+
Result table:
+-----------+------------+---------------------+
| player_id | event_date | games_played_so_far |
+-----------+------------+---------------------+
| 1 | 2016-03-01 | 5 |
| 1 | 2016-05-02 | 11 |
| 1 | 2017-06-25 | 12 |
| 3 | 2016-03-02 | 0 |
| 3 | 2018-07-03 | 5 |
+-----------+------------+---------------------+
For the player with id 1, 5 + 6 = 11 games played by 2016-05-02, and 5 + 6 + 1 = 12 games played by 2017-06-25.
For the player with id 3, 0 + 5 = 5 games played by 2018-07-03.
Note that for each player we only care about the days when the player logged in.

Solution

sql
#Solution- 1:
SELECT t1.player_id, t1.event_date, SUM(t2.games_played) as games_played_so_far
FROM Activity t1
JOIN Activity t2
ON t1.player_id = t2.player_id
WHERE t1.event_date >= t2.event_date
GROUP BY t1.player_id, t1.event_date;
#Solution- 2:
SELECT player_id, event_date,
SUM(games_played) OVER (PARTITION BY player_id ORDER BY event_date) AS games_played_so_far
FROM Activity;

550. Game Play Analysis IV | Medium | πŸ”’ LeetCode

Table: Activity

+--------------+---------+
| Column Name | Type |
+--------------+---------+
| player_id | int |
| device_id | int |
| event_date | date |
| games_played | int |
+--------------+---------+
(player_id, event_date) is the primary key of this table.
This table shows the activity of players of some game.
Each row is a record of a player who logged in and played a number of games (possibly 0) before logging out on some day using some device.

Write an SQL query that reports the fraction of players that logged in again on the day after the day they first logged in, rounded to 2 decimal places. In other words, you need to count the number of players that logged in for at least two consecutive days starting from their first login date, then divide that number by the total number of players.

The query result format is in the following example:

Activity table:
+-----------+-----------+------------+--------------+
| player_id | device_id | event_date | games_played |
+-----------+-----------+------------+--------------+
| 1 | 2 | 2016-03-01 | 5 |
| 1 | 2 | 2016-03-02 | 6 |
| 2 | 3 | 2017-06-25 | 1 |
| 3 | 1 | 2016-03-02 | 0 |
| 3 | 4 | 2018-07-03 | 5 |
+-----------+-----------+------------+--------------+
Result table:
+-----------+
| fraction |
+-----------+
| 0.33 |
+-----------+
Only the player with id 1 logged back in after the first day he had logged in so the answer is 1/3 = 0.33

Solution

sql
#Solution- 1:
SELECT ROUND(sum(CASE WHEN t1.event_date = t2.first_event+1 THEN 1 ELSE 0 END)/COUNT(DISTINCT t1.player_id), 2) AS fraction
FROM Activity t1
JOIN
(SELECT player_id, MIN(event_date) AS first_event
FROM Activity
GROUP BY player_id) t2
ON t1.player_id = t2.player_id;
#Solution- 2:
SELECT ROUND(COUNT(DISTINCT b.player_id)/COUNT(DISTINCT a.player_id),2) AS fraction
FROM
(SELECT player_id, MIN(event_date) AS event_date FROM Activity
GROUP BY player_id) a
LEFT JOIN Activity b
ON a.player_id = b.player_id AND a.event_date+1 = b.event_date;

569. Median Employee Salary | Hard | πŸ”’ LeetCode

The Employee table holds all employees. The employee table has three columns: Employee Id, Company Name, and Salary.

+-----+------------+--------+
|Id | Company | Salary |
+-----+------------+--------+
|1 | A | 2341 |
|2 | A | 341 |
|3 | A | 15 |
|4 | A | 15314 |
|5 | A | 451 |
|6 | A | 513 |
|7 | B | 15 |
|8 | B | 13 |
|9 | B | 1154 |
|10 | B | 1345 |
|11 | B | 1221 |
|12 | B | 234 |
|13 | C | 2345 |
|14 | C | 2645 |
|15 | C | 2645 |
|16 | C | 2652 |
|17 | C | 65 |
+-----+------------+--------+

Write a SQL query to find the median salary of each company. Bonus points if you can solve it without using any built-in SQL functions.

+-----+------------+--------+
|Id | Company | Salary |
+-----+------------+--------+
|5 | A | 451 |
|6 | A | 513 |
|12 | B | 234 |
|9 | B | 1154 |
|14 | C | 2645 |
+-----+------------+--------+

Solution

sql
SELECT t1.Id AS Id, t1.Company, t1.Salary
FROM Employee AS t1 JOIN Employee AS t2
ON t1.Company = t2.Company
GROUP BY t1.Id
HAVING abs(sum(CASE WHEN t2.Salary<t1.Salary THEN 1
WHEN t2.Salary>t1.Salary THEN -1
WHEN t2.Salary=t1.Salary AND t2.Id<t1.Id THEN 1
WHEN t2.Salary=t1.Salary AND t2.Id>t1.Id THEN -1
ELSE 0 END)) <= 1
ORDER BY t1.Company, t1.Salary, t1.Id

570. Managers with at Least 5 Direct Reports | Medium | πŸ”’ LeetCode

The Employee table holds all employees including their managers. Every employee has an Id, and there is also a column for the manager Id.

+------+----------+-----------+----------+
|Id |Name |Department |ManagerId |
+------+----------+-----------+----------+
|101 |John |A |null |
|102 |Dan |A |101 |
|103 |James |A |101 |
|104 |Amy |A |101 |
|105 |Anne |A |101 |
|106 |Ron |B |101 |
+------+----------+-----------+----------+

Given the Employee table, write a SQL query that finds out managers with at least 5 direct report. For the above table, your SQL query should return:

+-------+
| Name |
+-------+
| John |
+-------+

Note: No one would report to himself.

Solution

sql
SELECT Name
FROM Employee
WHERE id IN
(SELECT ManagerId
FROM Employee
GROUP BY ManagerId
HAVING COUNT(DISTINCT Id) >= 5)

571. Find Median Given Frequency of Numbers | πŸ”’ LeetCode

The Numbers table keeps the value of number and its frequency.

+----------+-------------+
| Number | Frequency |
+----------+-------------|
| 0 | 7 |
| 1 | 1 |
| 2 | 3 |
| 3 | 1 |
+----------+-------------+

In this table, the numbers are 0, 0, 0, 0, 0, 0, 0, 1, 2, 2, 2, 3, so the median is (0 + 0) / 2 = 0.

+--------+
| median |
+--------|
| 0.0000 |
+--------+

Write a query to find the median of all numbers and name the result as median.

Solution

sql
SELECT avg(t3.Number) as median
FROM Numbers as t3
JOIN
(SELECT t1.Number,
abs(SUM(CASE WHEN t1.Number>t2.Number THEN t2.Frequency ELSE 0 END) -
SUM(CASE WHEN t1.Number<t2.Number THEN t2.Frequency ELSE 0 END)) AS count_diff
FROM numbers AS t1, numbers AS t2
GROUP BY t1.Number) AS t4
ON t3.Number = t4.Number
WHERE t3.Frequency>=t4.count_diff

574. Winning Candidate | Medium | πŸ”’ LeetCode

Table: Candidate

+-----+---------+
| id | Name |
+-----+---------+
| 1 | A |
| 2 | B |
| 3 | C |
| 4 | D |
| 5 | E |
+-----+---------+

Table: Vote

+-----+--------------+
| id | CandidateId |
+-----+--------------+
| 1 | 2 |
| 2 | 4 |
| 3 | 3 |
| 4 | 2 |
| 5 | 5 |
+-----+--------------+

id is the auto-increment primary key, CandidateId is the id appeared in Candidate table. Write a sql to find the name of the winning candidate, the above example will return the winner B.

+------+
| Name |
+------+
| B |
+------+

Notes: You may assume there is no tie, in other words there will be at most one winning candidate.

Solution

sql
SELECT Name
FROM Candidate
WHERE id = (SELECT CandidateId
FROM Vote
GROUP BY CandidateId
ORDER BY COUNT(1) desc
LIMIT 1)
## Assumption: if we have two candidates with the same votes, we choose the one who get the first vote
# SELECT Name
# FROM Candidate JOIN
# (SELECT CandidateId
# FROM Vote
# GROUP BY CandidateId
# ORDER BY count(1) DESC
# LIMIT 1) AS t
# ON Candidate.id = t.CandidateId

577. Employee Bonus | Easy | πŸ”’ LeetCode

Select all employee’s name and bonus whose bonus is < 1000.

Table:Employee

+-------+--------+-----------+--------+
| empId | name | supervisor| salary |
+-------+--------+-----------+--------+
| 1 | John | 3 | 1000 |
| 2 | Dan | 3 | 2000 |
| 3 | Brad | null | 4000 |
| 4 | Thomas | 3 | 4000 |
+-------+--------+-----------+--------+
empId is the primary key column for this table.

Table: Bonus

+-------+-------+
| empId | bonus |
+-------+-------+
| 2 | 500 |
| 4 | 2000 |
+-------+-------+
empId is the primary key column for this table.

Example ouput:

+-------+-------+
| name | bonus |
+-------+-------+
| John | null |
| Dan | 500 |
| Brad | null |
+-------+-------+

Solution

sql
SELECT name, bonus
FROM Employee LEFT JOIN Bonus
ON Employee.empId = Bonus.empId
WHERE bonus<1000 OR bonus IS NULL;

578. Get Highest Answer Rate Question | Medium | πŸ”’ LeetCode

Get the highest answer rate question from a table survey_log with these columns: uid, action, question_id, answer_id, q_num, timestamp.

uid means user id; action has these kind of values: β€œshow”, β€œanswer”, β€œskip”; answer_id is not null when action column is β€œanswer”, while is null for β€œshow” and β€œskip”; q_num is the numeral order of the question in current session.

Write a sql query to identify the question which has the highest answer rate.

Example: Input:

+------+-----------+--------------+------------+-----------+------------+
| uid | action | question_id | answer_id | q_num | timestamp |
+------+-----------+--------------+------------+-----------+------------+
| 5 | show | 285 | null | 1 | 123 |
| 5 | answer | 285 | 124124 | 1 | 124 |
| 5 | show | 369 | null | 2 | 125 |
| 5 | skip | 369 | null | 2 | 126 |
+------+-----------+--------------+------------+-----------+------------+

Output:

+-------------+
| survey_log |
+-------------+
| 285 |
+-------------+

Explanation: question 285 has answer rate 1/1, while question 369 has 0/1 answer rate, so output 285.

Note: The highest answer rate meaning is: answer number’s ratio in show number in the same question.

Solution

sql
#Solution- 1::
SELECT question_id AS survey_log FROM
(SELECT question_id,
SUM(IF(action='show', 1, 0)) AS num_show,
SUM(IF(action='answer', 1, 0)) AS num_answer
FROM survey_log GROUP BY question_id) AS t
ORDER BY (num_answer/num_show) DESC LIMIT 1;
#Solution- 2:
SELECT question_id AS survey_log
FROM (SELECT question_id,
sum(CASE WHEN action='show' THEN 1 ELSE 0 END) AS show_count,
sum(CASE WHEN action='answer' THEN 1 ELSE 0 END) AS answer_count
FROM survey_log
GROUP BY question_id) AS t
ORDER BY answer_count/show_count DESC LIMIT 1;

579. Find Cumulative Salary of an Employee | Hard | πŸ”’ LeetCode

The Employee table holds the salary information in a year.

Write a SQL to get the cumulative sum of an employee’s salary over a period of 3 months but exclude the most recent month.

The result should be displayed by β€˜Id’ ascending, and then by β€˜Month’ descending.

Example Input

| Id | Month | Salary |
|----|-------|--------|
| 1 | 1 | 20 |
| 2 | 1 | 20 |
| 1 | 2 | 30 |
| 2 | 2 | 30 |
| 3 | 2 | 40 |
| 1 | 3 | 40 |
| 3 | 3 | 60 |
| 1 | 4 | 60 |
| 3 | 4 | 70 |

Output

| Id | Month | Salary |
|----|-------|--------|
| 1 | 3 | 90 |
| 1 | 2 | 50 |
| 1 | 1 | 20 |
| 2 | 1 | 20 |
| 3 | 3 | 100 |
| 3 | 2 | 40 |

Explanation Employee β€˜1’ has 3 salary records for the following 3 months except the most recent month β€˜4’: salary 40 for month β€˜3’, 30 for month β€˜2’ and 20 for month β€˜1’ So the cumulative sum of salary of this employee over 3 months is 90(40+30+20), 50(30+20) and 20 respectively.

| Id | Month | Salary |
|----|-------|--------|
| 1 | 3 | 90 |
| 1 | 2 | 50 |
| 1 | 1 | 20 |

Employee β€˜2’ only has one salary record (month β€˜1’) except its most recent month β€˜2’.

| Id | Month | Salary |
|----|-------|--------|
| 2 | 1 | 20 |

Employ β€˜3’ has two salary records except its most recent pay month β€˜4’: month β€˜3’ with 60 and month β€˜2’ with 40. So the cumulative salary is as following.

| Id | Month | Salary |
|----|-------|--------|
| 3 | 3 | 100 |
| 3 | 2 | 40 |

Solution

sql
SELECT
a.id,
a.month,
SUM(b.salary) Salary
FROM
Employee a JOIN Employee b ON
a.id = b.id AND
a.month - b.month >= 0 AND
a.month - b.month < 3
GROUP BY
a.id, a.month
HAVING
(a.id, a.month) NOT IN (SELECT id, MAX(month) FROM Employee GROUP BY id)
ORDER BY
a.id, a.month DESC

580. Count Student Number in Departments | Medium | πŸ”’ LeetCode

A university uses 2 data tables, student and department, to store data about its students and the departments associated with each major.

Write a query to print the respective department name and number of students majoring in each department for all departments in the department table (even ones with no current students).

Sort your results by descending number of students; if two or more departments have the same number of students, then sort those departments alphabetically by department name.

The student is described as follow:

| Column Name | Type |
|--------------|-----------|
| student_id | Integer |
| student_name | String |
| gender | Character |
| dept_id | Integer |

where student_id is the student’s ID number, student_name is the student’s name, gender is their gender, and dept_id is the department ID associated with their declared major.

And the department table is described as below:

| Column Name | Type |
|-------------|---------|
| dept_id | Integer |
| dept_name | String |

where dept_id is the department’s ID number and dept_name is the department name.

Here is an example input: student table:

| student_id | student_name | gender | dept_id |
|------------|--------------|--------|---------|
| 1 | Jack | M | 1 |
| 2 | Jane | F | 1 |
| 3 | Mark | M | 2 |

department table:

| dept_id | dept_name |
|---------|-------------|
| 1 | Engineering |
| 2 | Science |
| 3 | Law |

The Output should be:

| dept_name | student_number |
|-------------|----------------|
| Engineering | 2 |
| Science | 1 |
| Law | 0 |

Solution

sql
SELECT dept_name,
SUM(CASE WHEN student_id IS NULL THEN 0 ELSE 1 END) AS student_number
FROM department
LEFT JOIN student
ON department.dept_id = student.dept_id
GROUP BY department.dept_id
ORDER BY student_number DESC, dept_name

584. Find Customer Referee | Easy | πŸ”’ LeetCode

Given a table customer holding customers information and the referee.

+------+------+-----------+
| id | name | referee_id|
+------+------+-----------+
| 1 | Will | NULL |
| 2 | Jane | NULL |
| 3 | Alex | 2 |
| 4 | Bill | NULL |
| 5 | Zack | 1 |
| 6 | Mark | 2 |
+------+------+-----------+

Write a query to return the list of customers NOT referred by the person with id β€˜2’.

For the sample data above, the result is:

+------+
| name |
+------+
| Will |
| Jane |
| Bill |
| Zack |
+------+

Solution

sql
SELECT name
FROM customer
WHERE referee_id != '2' OR referee_id IS NULL;

585. Investments in 2016 | Medium | πŸ”’ LeetCode

Write a query to print the sum of all total investment values in 2016 (TIV_2016), to a scale of 2 decimal places, for all policy holders who meet the following criteria:

  1. Have the same TIV_2015 value as one or more other policyholders.
  2. Are not located in the same city as any other policyholder (i.e.: the (latitude, longitude) attribute pairs must be unique). Input Format: The insurance table is described as follows:
| Column Name | Type |
|-------------|---------------|
| PID | INTEGER(11) |
| TIV_2015 | NUMERIC(15,2) |
| TIV_2016 | NUMERIC(15,2) |
| LAT | NUMERIC(5,2) |
| LON | NUMERIC(5,2) |

where PID is the policyholder’s policy ID, TIV_2015 is the total investment value in 2015, TIV_2016 is the total investment value in 2016, LAT is the latitude of the policy holder’s city, and LON is the longitude of the policy holder’s city.

Sample Input

| PID | TIV_2015 | TIV_2016 | LAT | LON |
|-----|----------|----------|-----|-----|
| 1 | 10 | 5 | 10 | 10 |
| 2 | 20 | 20 | 20 | 20 |
| 3 | 10 | 30 | 20 | 20 |
| 4 | 10 | 40 | 40 | 40 |

Sample Output

| TIV_2016 |
|----------|
| 45.00 |

Explanation

The first record in the table, like the last record, meets both of the two criteria.
The TIV_2015 value '10' is as the same as the third and forth record, and its location unique.
The second record does not meet any of the two criteria. Its TIV_2015 is not like any other policyholders.
And its location is the same with the third record, which makes the third record fail, too.
So, the result is the sum of TIV_2016 of the first and last record, which is 45.

Solution

sql
SELECT SUM(TIV_2016) AS TIV_2016
FROM insurance
WHERE CONCAT(LAT, ',', LON)
IN (SELECT CONCAT(LAT, ',', LON)
FROM insurance
GROUP BY LAT, LON
HAVING COUNT(1) = 1)
AND TIV_2015 in
(SELECT TIV_2015
FROM insurance
GROUP BY TIV_2015
HAVING COUNT(1)>1)

586. Customer Placing the Largest Number of Orders | Easy | πŸ”’ LeetCode

Query the customer_number from the orders table for the customer who has placed the largest number of orders.

It is guaranteed that exactly one customer will have placed more orders than any other customer.

The orders table is defined as follows:

| Column | Type |
|-------------------|-----------|
| order_number (PK) | int |
| customer_number | int |
| order_date | date |
| required_date | date |
| shipped_date | date |
| status | char(15) |
| comment | char(200) |

Sample Input

| order_number | customer_number | order_date | required_date | shipped_date | status | comment |
|--------------|-----------------|------------|---------------|--------------|--------|---------|
| 1 | 1 | 2017-04-09 | 2017-04-13 | 2017-04-12 | Closed | |
| 2 | 2 | 2017-04-15 | 2017-04-20 | 2017-04-18 | Closed | |
| 3 | 3 | 2017-04-16 | 2017-04-25 | 2017-04-20 | Closed | |
| 4 | 3 | 2017-04-18 | 2017-04-28 | 2017-04-25 | Closed | |

Sample Output

| customer_number |
|-----------------|
| 3 |

Explanation

The customer with number '3' has two orders, which is greater than either customer '1' or '2' because each of them only has one order.
So the result is customer_number '3'.

Solution

sql
# assume: only one match
SELECT customer_number FROM orders
GROUP BY customer_number
ORDER BY COUNT(1) DESC
LIMIT 1
## assume: multiple matches
## 1 1
## 2 1
## 3 1
##
## 1 1 1 1
## 1 1 2 1
## 1 1 3 1
##
## SELECT t1.customer_number
## FROM (SELECT customer_number, COUNT(1) AS count
## FROM orders GROUP BY customer_number) AS t1,
## (SELECT customer_number, COUNT(1) AS count
## FROM orders GROUP BY customer_number) AS t2
## GROUP BY t1.customer_number
## HAVING max(t1.count) = max(t2.count)

595. Big Countries | Easy | LeetCode

There is a table World

+-----------------+------------+------------+--------------+---------------+
| name | continent | area | population | gdp |
+-----------------+------------+------------+--------------+---------------+
| Afghanistan | Asia | 652230 | 25500100 | 20343000 |
| Albania | Europe | 28748 | 2831741 | 12960000 |
| Algeria | Africa | 2381741 | 37100000 | 188681000 |
| Andorra | Europe | 468 | 78115 | 3712000 |
| Angola | Africa | 1246700 | 20609294 | 100990000 |
+-----------------+------------+------------+--------------+---------------+

A country is big if it has an area of bigger than 3 million square km or a population of more than 25 million.

Write a SQL solution to output big countries' name, population and area.

For example, according to the above table, we should output:

+--------------+-------------+--------------+
| name | population | area |
+--------------+-------------+--------------+
| Afghanistan | 25500100 | 652230 |
| Algeria | 37100000 | 2381741 |
+--------------+-------------+--------------+

Solution

sql
SELECT name, population, area
FROM World
WHERE area >= 3000000 OR population > 25000000;

596. Classes More Than 5 Students | Easy | LeetCode

There is a table courses with columns: student and class

Please list out all classes which have more than or equal to 5 students.

For example, the table:

+---------+------------+
| student | class |
+---------+------------+
| A | Math |
| B | English |
| C | Math |
| D | Biology |
| E | Math |
| F | Computer |
| G | Math |
| H | Math |
| I | Math |
+---------+------------+

Should output:

+---------+
| class |
+---------+
| Math |
+---------+

Solution

sql
SELECT class
FROM courses
GROUP BY class
HAVING count(DISTINCT Student)>=5;

597. Friend Requests I: Overall Acceptance Rate | Easy | πŸ”’ LeetCode

In social network like Facebook or Twitter, people send friend requests and accept others’ requests as well. Now given two tables as below: Table: friend_request

| sender_id | send_to_id |request_date|
|-----------|------------|------------|
| 1 | 2 | 2016_06-01 |
| 1 | 3 | 2016_06-01 |
| 1 | 4 | 2016_06-01 |
| 2 | 3 | 2016_06-02 |
| 3 | 4 | 2016-06-09 |

Table: request_accepted

| requester_id | accepter_id |accept_date |
|--------------|-------------|------------|
| 1 | 2 | 2016_06-03 |
| 1 | 3 | 2016-06-08 |
| 2 | 3 | 2016-06-08 |
| 3 | 4 | 2016-06-09 |
| 3 | 4 | 2016-06-10 |

Write a query to find the overall acceptance rate of requests rounded to 2 decimals, which is the number of acceptance divide the number of requests. For the sample data above, your query should return the following result.

|accept_rate|
|-----------|
| 0.80|

Note:

The accepted requests are not necessarily from the table friend_request. In this case, you just need to simply count the total accepted requests (no matter whether they are in the original requests), and divide it by the number of requests to get the acceptance rate. It is possible that a sender sends multiple requests to the same receiver, and a request could be accepted more than once. In this case, the β€˜duplicated’ requests or acceptances are only counted once. If there is no requests at all, you should return 0.00 as the accept_rate. Explanation: There are 4 unique accepted requests, and there are 5 requests in total. So the rate is 0.80.

Follow-up:

Can you write a query to return the accept rate but for every month? How about the cumulative accept rate for every day?

Solution

sql
SELECT IFNULL((round(accepts/requests, 2)), 0.0) AS accept_rate
FROM
(SELECT count(DISTINCT sender_id, send_to_id) AS requests FROM friend_request) AS t1,
(SELECT count(DISTINCT requester_id, accepter_id) AS accepts FROM request_accepted) AS t2

601. Human Traffic of Stadium | Hard | LeetCode

Table: Stadium

+---------------+---------+
| Column Name | Type |
+---------------+---------+
| id | int |
| visit_date | date |
| people | int |
+---------------+---------+

visit_date is the primary key for this table. Each row of this table contains the visit date and visit id to the stadium with the number of people during the visit. No two rows will have the same visit_date, and as the id increases, the dates increase as well.

Write an SQL query to display the records with three or more rows with consecutive id's, and the number of people is greater than or equal to 100 for each.

Return the result table ordered by visit_date in ascending order.

The query result format is in the following example.

Stadium table:
+------+------------+-----------+
| id | visit_date | people |
+------+------------+-----------+
| 1 | 2017-01-01 | 10 |
| 2 | 2017-01-02 | 109 |
| 3 | 2017-01-03 | 150 |
| 4 | 2017-01-04 | 99 |
| 5 | 2017-01-05 | 145 |
| 6 | 2017-01-06 | 1455 |
| 7 | 2017-01-07 | 199 |
| 8 | 2017-01-09 | 188 |
+------+------------+-----------+
Result table:
+------+------------+-----------+
| id | visit_date | people |
+------+------------+-----------+
| 5 | 2017-01-05 | 145 |
| 6 | 2017-01-06 | 1455 |
| 7 | 2017-01-07 | 199 |
| 8 | 2017-01-09 | 188 |
+------+------------+-----------+
The four rows with ids 5, 6, 7, and 8 have consecutive ids and each of them has >= 100 people attended. Note that row 8 was included even though the visit_date was not the next day after row 7.
The rows with ids 2 and 3 are not included because we need at least three consecutive ids.

Solution

sql
SELECT DISTINCT s1.*
FROM Stadium s1 JOIN Stadium s2 JOIN Stadium s3
ON (s1.id = s2.id-1 AND s1.id = s3.id-2) OR
(s1.id = s2.id+1 AND s1.id = s3.id-1) OR
(s1.id = s2.id+1 AND s1.id = s3.id+2)
WHERE s1.people >= 100 AND s2.people >= 100 AND s3.people>=100
ORDER BY visit_date

602. Friend Requests II: Who Has the Most Friends | Medium | πŸ”’ LeetCode

In social network like Facebook or Twitter, people send friend requests and accept others’ requests as well. Table request_accepted holds the data of friend acceptance, while requester_id and accepter_id both are the id of a person.

| requester_id | accepter_id | accept_date|
|--------------|-------------|------------|
| 1 | 2 | 2016_06-03 |
| 1 | 3 | 2016-06-08 |
| 2 | 3 | 2016-06-08 |
| 3 | 4 | 2016-06-09 |

Write a query to find the the people who has most friends and the most friends number. For the sample data above, the result is:

| id | num |
|----|-----|
| 3 | 3 |

Note:

It is guaranteed there is only 1 people having the most friends. The friend request could only been accepted once, which mean there is no multiple records with the same requester_id and accepter_id value. Explanation: The person with id β€˜3’ is a friend of people β€˜1’, β€˜2’ and β€˜4’, so he has 3 friends in total, which is the most number than any others.

Follow-up: In the real world, multiple people could have the same most number of friends, can you find all these people in this case?

sql
SELECT t.id, sum(t.num) AS num
FROM (
(SELECT requester_id AS id, COUNT(1) AS num
FROM request_accepted
GROUP BY requester_id)
union all
(SELECT accepter_id AS id, COUNT(1) AS num
FROM request_accepted
GROUP BY accepter_id)) AS t
GROUP BY t.id
ORDER BY num DESC
LIMIT 1;

603. Consecutive Available Seats | Easy | πŸ”’ LeetCode

Several friends at a cinema ticket office would like to reserve consecutive available seats. Can you help to query all the consecutive available seats order by the seat_id using the following cinema table?

| seat_id | free |
|---------|------|
| 1 | 1 |
| 2 | 0 |
| 3 | 1 |
| 4 | 1 |
| 5 | 1 |

Your query should return the following result for the sample case above.

| seat_id |
|---------|
| 3 |
| 4 |
| 5 |

Note:

The seat_id is an auto increment int, and free is bool (β€˜1’ means free, and β€˜0’ means occupied.). Consecutive available seats are more than 2(inclusive) seats consecutively available.

Solution

sql
SELECT DISTINCT t1.seat_id
FROM cinema AS t1 JOIN cinema AS t2
ON abs(t1.seat_id-t2.seat_id)=1
WHERE t1.free='1' AND t2.free='1'
ORDER BY t1.seat_id

607.Sales Person | Easy | πŸ”’ LeetCode

Description

Given three tables: salesperson, company, orders. Output all the names in the table salesperson, who didn’t have sales to company β€˜RED’.

Example Input

Table: salesperson

+----------+------+--------+-----------------+-----------+
| sales_id | name | salary | commission_rate | hire_date |
+----------+------+--------+-----------------+-----------+
| 1 | John | 100000 | 6 | 4/1/2006 |
| 2 | Amy | 120000 | 5 | 5/1/2010 |
| 3 | Mark | 65000 | 12 | 12/25/2008|
| 4 | Pam | 25000 | 25 | 1/1/2005 |
| 5 | Alex | 50000 | 10 | 2/3/2007 |
+----------+------+--------+-----------------+-----------+

The table salesperson holds the salesperson information. Every salesperson has a sales_id and a name. Table: company

+---------+--------+------------+
| com_id | name | city |
+---------+--------+------------+
| 1 | RED | Boston |
| 2 | ORANGE | New York |
| 3 | YELLOW | Boston |
| 4 | GREEN | Austin |
+---------+--------+------------+

The table company holds the company information. Every company has a com_id and a name. Table: orders

+----------+----------+---------+----------+--------+
| order_id | date | com_id | sales_id | amount |
+----------+----------+---------+----------+--------+
| 1 | 1/1/2014 | 3 | 4 | 100000 |
| 2 | 2/1/2014 | 4 | 5 | 5000 |
| 3 | 3/1/2014 | 1 | 1 | 50000 |
| 4 | 4/1/2014 | 1 | 4 | 25000 |
+----------+----------+---------+----------+--------+

The table orders holds the sales record information, salesperson and customer company are represented by sales_id and com_id. output

+------+
| name |
+------+
| Amy |
| Mark |
| Alex |
+------+

Explanation

According to order β€˜3’ and β€˜4’ in table orders, it is easy to tell only salesperson β€˜John’ and β€˜Alex’ have sales to company β€˜RED’, so we need to output all the other names in table salesperson.

Solution

sql
SELECT name
FROM salesperson
WHERE name NOT IN
(SELECT DISTINCT salesperson.name
FROM salesperson, orders, company
WHERE company.name = 'RED'
AND salesperson.sales_id = orders.sales_id
AND orders.com_id = company.com_id)

608. Tree Node | Medium | πŸ”’ LeetCode

Given a table tree, id is identifier of the tree node and p_id is its parent node’s id.

+----+------+
| id | p_id |
+----+------+
| 1 | null |
| 2 | 1 |
| 3 | 1 |
| 4 | 2 |
| 5 | 2 |
+----+------+

Each node in the tree can be one of three types:

Leaf: if the node is a leaf node. Root: if the node is the root of the tree. Inner: If the node is neither a leaf node nor a root node. Write a query to print the node id and the type of the node. Sort your output by the node id. The result for the above sample is:

+----+------+
| id | Type |
+----+------+
| 1 | Root |
| 2 | Inner|
| 3 | Leaf |
| 4 | Leaf |
| 5 | Leaf |
+----+------+

Explanation

Node β€˜1’ is root node, because its parent node is NULL and it has child node β€˜2’ and β€˜3’. Node β€˜2’ is inner node, because it has parent node β€˜1’ and child node β€˜4’ and β€˜5’. Node β€˜3’, β€˜4’ and β€˜5’ is Leaf node, because they have parent node and they don’t have child node. And here is the image of the sample tree as below:

1
/ \
2 3
/ \
4 5

Note

If there is only one node on the tree, you only need to output its root attributes.

Solution

sql
## Basic Ideas: LEFT JOIN
# In tree, each node can only one parent or no parent
## | id | p_id | id (child) |
## |----+------+------------|
## | 1 | null | 1 |
## | 1 | null | 2 |
## | 2 | 1 | 4 |
## | 2 | 1 | 5 |
## | 3 | 1 | null |
## | 4 | 2 | null |
## | 5 | 2 | null |
SELECT t1.id,
CASE
WHEN ISNULL(t1.p_id) THEN 'Root'
WHEN ISNULL(MAX(t2.id)) THEN 'Leaf'
ELSE 'Inner'
END AS Type
FROM tree AS t1 LEFT JOIN tree AS t2
ON t1.id = t2.p_id
GROUP BY t1.id, t1.p_id

610. Triangle Judgement | Easy | πŸ”’ LeetCode

A pupil Tim gets homework to identify whether three line segments could possibly form a triangle. However, this assignment is very heavy because there are hundreds of records to calculate. Could you help Tim by writing a query to judge whether these three sides can form a triangle, assuming table triangle holds the length of the three sides x, y and z.

| x | y | z |
|----|----|----|
| 13 | 15 | 30 |
| 10 | 20 | 15 |

For the sample data above, your query should return the follow result:

| x | y | z | triangle |
|----|----|----|----------|
| 13 | 15 | 30 | No |
| 10 | 20 | 15 | Yes |

Solution

sql
SELECT x, y, z,
CASE
WHEN x+y>z AND y+z>x AND x+z>y THEN 'Yes'
ELSE 'No'
END AS triangle
FROM triangle

612. Shortest Distance in a Plane | Medium | πŸ”’ LeetCode

Table point_2d holds the coordinates (x,y) of some unique points (more than two) in a plane. Write a query to find the shortest distance between these points rounded to 2 decimals.

| x | y |
|----|----|
| -1 | -1 |
| 0 | 0 |
| -1 | -2 |

The shortest distance is 1.00 from point (-1,-1) to (-1,2). So the output should be:

| shortest |
|----------|
| 1.00 |

Note: The longest distance among all the points are less than 10000.

Solution

sql
SELECT ROUND(MIN(SQRT((t1.x-t2.x)*(t1.x-t2.x) + (t1.y-t2.y)*(t1.y-t2.y))), 2) as shortest
FROM point_2d AS t1, point_2d AS t2
WHERE t1.x!=t2.x OR t1.y!=t2.y
# SELECT ROUND(SQRT((t1.x-t2.x)*(t1.x-t2.x) + (t1.y-t2.y)*(t1.y-t2.y)), 2) AS shortest
# FROM point_2d AS t1, point_2d AS t2
# WHERE t1.x!=t2.x OR t1.y!=t2.y
# ORDER BY shortest ASC
# LIMIT 1

613. Shortest Distance in a Line | Easy | πŸ”’ LeetCode

Table point holds the x coordinate of some points on x-axis in a plane, which are all integers. Write a query to find the shortest distance between two points in these points.

| x |
|-----|
| -1 |
| 0 |
| 2 |

The shortest distance is β€˜1’ obviously, which is from point β€˜-1’ to β€˜0’. So the output is as below:

| shortest|
|---------|
| 1 |

Note: Every point is unique, which means there is no duplicates in table point.

Follow-up: What if all these points have an id and are arranged from the left most to the right most of x axis?

Solution

sql
SELECT t1.x-t2.x AS shortest
FROM point AS t1 JOIN point AS t2
WHERE t1.x>t2.x
ORDER BY (t1.x-t2.x) ASC
LIMIT 1

614. Second Degree Follower | Medium | πŸ”’ LeetCode

In facebook, there is a follow table with two columns: followee, follower.

Please write a sql query to get the amount of each follower’s follower if he/she has one.

For example:

+-------------+------------+
| followee | follower |
+-------------+------------+
| A | B |
| B | C |
| B | D |
| D | E |
+-------------+------------+

should output:

+-------------+------------+
| follower | num |
+-------------+------------+
| B | 2 |
| D | 1 |
+-------------+------------+

Explanation: Both B and D exist in the follower list, when as a followee, B’s follower is C and D, and D’s follower is E. A does not exist in follower list.

Note: Followee would not follow himself/herself in all cases. Please display the result in follower’s alphabet order.

Solution

sql
## Explain the business logic
## A follows B. Then A is follwer, B is followee
## What are second degree followers?
## A follows B, and B follows C.
## Then A is the second degree followers of C
SELECT f1.follower, COUNT(DISTINCT f2.follower) AS num
FROM follow AS f1 JOIN follow AS f2
ON f1.follower = f2.followee
GROUP BY f1.follower;

615. Average Salary: Departments VS Company | Hard | πŸ”’ LeetCode

Given two tables as below, write a query to display the comparison result (higher/lower/same) of the average salary of employees in a department to the company’s average salary. Table: salary

| id | employee_id | amount | pay_date |
|----|-------------|--------|------------|
| 1 | 1 | 9000 | 2017-03-31 |
| 2 | 2 | 6000 | 2017-03-31 |
| 3 | 3 | 10000 | 2017-03-31 |
| 4 | 1 | 7000 | 2017-02-28 |
| 5 | 2 | 6000 | 2017-02-28 |
| 6 | 3 | 8000 | 2017-02-28 |

The employee_id column refers to the employee_id in the following table employee.

| employee_id | department_id |
|-------------|---------------|
| 1 | 1 |
| 2 | 2 |
| 3 | 2 |

So for the sample data above, the result is:

| pay_month | department_id | comparison |
|-----------|---------------|-------------|
| 2017-03 | 1 | higher |
| 2017-03 | 2 | lower |
| 2017-02 | 1 | same |
| 2017-02 | 2 | same |

Explanation In March, the company’s average salary is (9000+6000+10000)/3 = 8333.33… The average salary for department β€˜1’ is 9000, which is the salary of employee_id β€˜1’ since there is only one employee in this department. So the comparison result is β€˜higher’ since 9000 > 8333.33 obviously. The average salary of department β€˜2’ is (6000 + 10000)/2 = 8000, which is the average of employee_id β€˜2’ and β€˜3’. So the comparison result is β€˜lower’ since 8000 < 8333.33. With he same formula for the average salary comparison in February, the result is β€˜same’ since both the department β€˜1’ and β€˜2’ have the same average salary with the company, which is 7000.

Solution

sql
SELECT t1.pay_month, t1.department_id,
(CASE WHEN t1.amount = t2.amount THEN 'same'
WHEN t1.amount > t2.amount THEN 'higher'
WHEN t1.amount < t2.amount THEN 'lower' END) AS comparison
FROM
(SELECT left(pay_date, 7) AS pay_month, department_id, avg(amount) AS amount
FROM salary JOIN employee
ON salary.employee_id = employee.employee_id
GROUP BY pay_month, department_id
ORDER BY pay_month DESC, department_id) AS t1
JOIN
(SELECT left(pay_date, 7) AS pay_month, avg(amount) AS amount
FROM salary JOIN employee
ON salary.employee_id = employee.employee_id
GROUP BY pay_month) AS t2
ON t1.pay_month = t2.pay_month

618. Students Report By Geography | Hard | πŸ”’ LeetCode

A U.S graduate school has students from Asia, Europe and America. The students’ location information are stored in table student as below.

| name | continent |
|--------|-----------|
| Jack | America |
| Pascal | Europe |
| Xi | Asia |
| Jane | America |

Pivot the continent column in this table so that each name is sorted alphabetically and displayed underneath its corresponding continent. The output headers should be America, Asia and Europe respectively. It is guaranteed that the student number from America is no less than either Asia or Europe. For the sample input, the output is:

| America | Asia | Europe |
|---------|------|--------|
| Jack | Xi | Pascal |
| Jane | | |

Follow-up: If it is unknown which continent has the most students, can you write a query to generate the student report?

Solution

sql
SELECT t1.name AS America, t2.name AS Asia, t3.name AS Europe
FROM
(SELECT (@cnt1 := @cnt1 + 1) AS id, name
FROM student
CROSS JOIN (SELECT @cnt1 := 0) AS dummy
WHERE continent='America'
ORDER BY name) AS t1
LEFT JOIN
(SELECT (@cnt2 := @cnt2 + 1) AS id, name
FROM student
CROSS JOIN (SELECT @cnt2 := 0) AS dummy
WHERE continent='Asia'
ORDER BY name) AS t2
ON t1.id = t2.id
LEFT JOIN
(SELECT (@cnt3 := @cnt3 + 1) AS id, name
FROM student
CROSS JOIN (SELECT @cnt3 := 0) AS dummy
WHERE continent='Europe'
ORDER BY name) AS t3
ON t1.id = t3.id

619. Biggest Single Number | Easy | πŸ”’ LeetCode

Table number contains many numbers in column num including duplicated ones. Can you write a SQL query to find the biggest number, which only appears once.

+---+
|num|
+---+
| 8 |
| 8 |
| 3 |
| 3 |
| 1 |
| 4 |
| 5 |
| 6 |

For the sample data above, your query should return the following result:

+---+
|num|
+---+
| 6 |

Note: If there is no such number, just output null.

Solution

sql
SELECT IFNULL((
SELECT num
FROM number
GROUP BY num
HAVING count(1) = 1
ORDER BY num DESC
LIMIT 0, 1), NULL) AS num

620. Not Boring Movies | Easy | LeetCode

X city opened a new cinema, many people would like to go to this cinema. The cinema also gives out a poster indicating the movies’ ratings and descriptions. Please write a SQL query to output movies with an odd numbered ID and a description that is not 'boring'. Order the result by rating.

For example, table cinema:

+---------+-----------+--------------+-----------+
| id | movie | description | rating |
+---------+-----------+--------------+-----------+
| 1 | War | great 3D | 8.9 |
| 2 | Science | fiction | 8.5 |
| 3 | irish | boring | 6.2 |
| 4 | Ice song | Fantacy | 8.6 |
| 5 | House card| Interesting| 9.1 |
+---------+-----------+--------------+-----------+

For the example above, the output should be:

+---------+-----------+--------------+-----------+
| id | movie | description | rating |
+---------+-----------+--------------+-----------+
| 5 | House card| Interesting| 9.1 |
| 1 | War | great 3D | 8.9 |
+---------+-----------+--------------+-----------+

Solution

sql
SELECT *
FROM Cinema
WHERE description <> 'boring' AND ID % 2 = 1
ORDER BY rating DESC;

626. Exchange Seats | Medium | LeetCode

Mary is a teacher in a middle school and she has a table seat storing students' names and their corresponding seat ids.

The column id is continuous increment.

Mary wants to change seats for the adjacent students.

Can you write a SQL query to output the result for Mary?

+---------+---------+
| id | student |
+---------+---------+
| 1 | Abbot |
| 2 | Doris |
| 3 | Emerson |
| 4 | Green |
| 5 | Jeames |
+---------+---------+

For the sample input, the output is:

+---------+---------+
| id | student |
+---------+---------+
| 1 | Doris |
| 2 | Abbot |
| 3 | Green |
| 4 | Emerson |
| 5 | Jeames |
+---------+---------+

Note:
If the number of students is odd, there is no need to change the last one's seat.

Solution

sql
SELECT
IF(id<(SELECT MAX(id) FROM seat),IF(id%2=0,id-1, id+1),IF(id%2=0, id-1, id)) AS id, student
FROM seat
ORDER BY id;

627. Swap Salary | LeetCode

Table: Salary

+-------------+----------+
| Column Name | Type |
+-------------+----------+
| id | int |
| name | varchar |
| sex | ENUM |
| salary | int |
+-------------+----------+
id is the primary key for this table.
The sex column is ENUM value of type ('m', 'f').
The table contains information about an employee.

Write an SQL query to swap all 'f' and 'm' values (i.e., change all 'f' values to 'm' and vice versa) with a single update statement and no intermediate temp table(s).

Note that you must write a single update statement, DO NOT write any select statement for this problem.

The query result format is in the following example:

Salary table:
+----+------+-----+--------+
| id | name | sex | salary |
+----+------+-----+--------+
| 1 | A | m | 2500 |
| 2 | B | f | 1500 |
| 3 | C | m | 5500 |
| 4 | D | f | 500 |
+----+------+-----+--------+
Result table:
+----+------+-----+--------+
| id | name | sex | salary |
+----+------+-----+--------+
| 1 | A | f | 2500 |
| 2 | B | m | 1500 |
| 3 | C | f | 5500 |
| 4 | D | m | 500 |
+----+------+-----+--------+
(1, A) and (2, C) were changed from 'm' to 'f'.
(2, B) and (4, D) were changed from 'f' to 'm'.

Solution

sql
# With IF
UPDATE Salary SET sex = IF(sex='m', 'f', 'm')
# With CASE
UPDATE Salary SET sex = CASE WHEN sex='m' THEN 'f' ELSE 'm' END

1045. Customers Who Bought All Products | Medium | πŸ”’ LeetCode

Table: Customer

+-------------+---------+
| Column Name | Type |
+-------------+---------+
| customer_id | int |
| product_key | int |
+-------------+---------+

product_key is a foreign key to Product table. Table: Product

+-------------+---------+
| Column Name | Type |
+-------------+---------+
| product_key | int |
+-------------+---------+
product_key is the primary key column for this table.

Write an SQL query for a report that provides the customer ids from the Customer table that bought all the products in the Product table.

For example:

Customer table:
+-------------+-------------+
| customer_id | product_key |
+-------------+-------------+
| 1 | 5 |
| 2 | 6 |
| 3 | 5 |
| 3 | 6 |
| 1 | 6 |
+-------------+-------------+
Product table:
+-------------+
| product_key |
+-------------+
| 5 |
| 6 |
+-------------+
Result table:
+-------------+
| customer_id |
+-------------+
| 1 |
| 3 |
+-------------+
The customers who bought all the products (5 and 6) are customers with id 1 and 3.

Solution

sql
SELECT customer_id
FROM Customer
GROUP NY customer_id
HAVING count(DISTINCT product_key) = (
SELECT count(1)
FROM Product)

1050. Actors and Directors Who Cooperated At Least Three Times | Easy | πŸ”’ LeetCode

Table: ActorDirector

+-------------+---------+
| Column Name | Type |
+-------------+---------+
| actor_id | int |
| director_id | int |
| timestamp | int |
+-------------+---------+
timestamp is the primary key column for this table.

Write a SQL query for a report that provides the pairs (actor_id, director_id) where the actor have cooperated with the director at least 3 times.

Example:

ActorDirector table:
+-------------+-------------+-------------+
| actor_id | director_id | timestamp |
+-------------+-------------+-------------+
| 1 | 1 | 0 |
| 1 | 1 | 1 |
| 1 | 1 | 2 |
| 1 | 2 | 3 |
| 1 | 2 | 4 |
| 2 | 1 | 5 |
| 2 | 1 | 6 |
+-------------+-------------+-------------+
Result table:
+-------------+-------------+
| actor_id | director_id |
+-------------+-------------+
| 1 | 1 |
+-------------+-------------+
The only pair is (1, 1) where they cooperated exactly 3 times.

Solution

sql
SELECT actor_id, director_id
FROM ActorDirector
GROUP BY actor_id, director_id
HAVING COUNT(1)>=3

1068. Product Sales Analysis I | Easy | πŸ”’ LeetCode

Table: Sales

+-------------+-------+
| Column Name | Type |
+-------------+-------+
| sale_id | int |
| product_id | int |
| year | int |
| quantity | int |
| price | int |
+-------------+-------+
(sale_id, year) is the primary key of this table.
product_id is a foreign key to Product table.
Note that the price is per unit.

Table: Product

+--------------+---------+
| Column Name | Type |
+--------------+---------+
| product_id | int |
| product_name | varchar |
+--------------+---------+
product_id is the primary key of this table.

Write an SQL query that reports all product names of the products in the Sales table along with their selling year and price.

For example:

Sales table:
+---------+------------+------+----------+-------+
| sale_id | product_id | year | quantity | price |
+---------+------------+------+----------+-------+
| 1 | 100 | 2008 | 10 | 5000 |
| 2 | 100 | 2009 | 12 | 5000 |
| 7 | 200 | 2011 | 15 | 9000 |
+---------+------------+------+----------+-------+
Product table:
+------------+--------------+
| product_id | product_name |
+------------+--------------+
| 100 | Nokia |
| 200 | Apple |
| 300 | Samsung |
+------------+--------------+
Result table:
+--------------+-------+-------+
| product_name | year | price |
+--------------+-------+-------+
| Nokia | 2008 | 5000 |
| Nokia | 2009 | 5000 |
| Apple | 2011 | 9000 |
+--------------+-------+-------+

Solution

sql
SELECT product_name, year, price
FROM Sales JOIN Product
ON Product.product_id = Sales.product_id

1069. Product Sales Analysis II | Easy | πŸ”’ LeetCode

Table: Sales

+-------------+-------+
| Column Name | Type |
+-------------+-------+
| sale_id | int |
| product_id | int |
| year | int |
| quantity | int |
| price | int |
+-------------+-------+
sale_id is the primary key of this table.
product_id is a foreign key to Product table.
Note that the price is per unit.

Table: Product

+--------------+---------+
| Column Name | Type |
+--------------+---------+
| product_id | int |
| product_name | varchar |
+--------------+---------+
product_id is the primary key of this table.

Write an SQL query that reports the total quantity sold for every product id.

The query result format is in the following example:

Sales table:
+---------+------------+------+----------+-------+
| sale_id | product_id | year | quantity | price |
+---------+------------+------+----------+-------+
| 1 | 100 | 2008 | 10 | 5000 |
| 2 | 100 | 2009 | 12 | 5000 |
| 7 | 200 | 2011 | 15 | 9000 |
+---------+------------+------+----------+-------+
Product table:
+------------+--------------+
| product_id | product_name |
+------------+--------------+
| 100 | Nokia |
| 200 | Apple |
| 300 | Samsung |
+------------+--------------+
Result table:
+--------------+----------------+
| product_id | total_quantity |
+--------------+----------------+
| 100 | 22 |
| 200 | 15 |
+--------------+----------------+

Solution

sql
SELECT product_id, sum(quantity) AS total_quantity
FROM Sales
GROUP BY product_id;

1070. Product Sales Analysis III | Medium | πŸ”’ LeetCode

Table: Sales

+-------------+-------+
| Column Name | Type |
+-------------+-------+
| sale_id | int |
| product_id | int |
| year | int |
| quantity | int |
| price | int |
+-------------+-------+
sale_id is the primary key of this table.
product_id is a foreign key to Product table.
Note that the price is per unit.

Table: Product

+--------------+---------+
| Column Name | Type |
+--------------+---------+
| product_id | int |
| product_name | varchar |
+--------------+---------+
product_id is the primary key of this table.

Write an SQL query that selects the product id, year, quantity, and price for the first year of every product sold.

The query result format is in the following example:

Sales table:
+---------+------------+------+----------+-------+
| sale_id | product_id | year | quantity | price |
+---------+------------+------+----------+-------+
| 1 | 100 | 2008 | 10 | 5000 |
| 2 | 100 | 2009 | 12 | 5000 |
| 7 | 200 | 2011 | 15 | 9000 |
+---------+------------+------+----------+-------+
Product table:
+------------+--------------+
| product_id | product_name |
+------------+--------------+
| 100 | Nokia |
| 200 | Apple |
| 300 | Samsung |
+------------+--------------+
Result table:
+------------+------------+----------+-------+
| product_id | first_year | quantity | price |
+------------+------------+----------+-------+
| 100 | 2008 | 10 | 5000 |
| 200 | 2011 | 15 | 9000 |
+------------+------------+----------+-------+

Solution

sql
SELECT
product_id,
year first_year,
quantity,
price
FROM Sales
WHERE (product_id, year) IN (SELECT product_id, MIN(year)
FROM Sales
GROUP BY product_id)

1075. Project Employees I | Easy | πŸ”’ LeetCode

Table: Project

+-------------+---------+
| Column Name | Type |
+-------------+---------+
| project_id | int |
| employee_id | int |
+-------------+---------+
(project_id, employee_id) is the primary key of this table.
employee_id is a foreign key to Employee table.

Table: Employee

+------------------+---------+
| Column Name | Type |
+------------------+---------+
| employee_id | int |
| name | varchar |
| experience_years | int |
+------------------+---------+
employee_id is the primary key of this table.

Write an SQL query that reports the average experience years of all the employees for each project, rounded to 2 digits.

The query result format is in the following example:

Project table:
+-------------+-------------+
| project_id | employee_id |
+-------------+-------------+
| 1 | 1 |
| 1 | 2 |
| 1 | 3 |
| 2 | 1 |
| 2 | 4 |
+-------------+-------------+
Employee table:
+-------------+--------+------------------+
| employee_id | name | experience_years |
+-------------+--------+------------------+
| 1 | Khaled | 3 |
| 2 | Ali | 2 |
| 3 | John | 1 |
| 4 | Doe | 2 |
+-------------+--------+------------------+
Result table:
+-------------+---------------+
| project_id | average_years |
+-------------+---------------+
| 1 | 2.00 |
| 2 | 2.50 |
+-------------+---------------+
The average experience years for the first project is (3 + 2 + 1) / 3 = 2.00 and for the second project is (3 + 2) / 2 = 2.50

Solution

sql
SELECT
p.project_id,
ROUND(AVG(e.experience_years),2) average_years
FROM
Project p JOIN Employee e ON
p.employee_id = e.employee_id
GROUP BY
p.project_id

1076. Project Employees II | Easy | πŸ”’ LeetCode

Table: Project

+-------------+---------+
| Column Name | Type |
+-------------+---------+
| project_id | int |
| employee_id | int |
+-------------+---------+
(project_id, employee_id) is the primary key of this table.
employee_id is a foreign key to Employee table.

Table: Employee

+------------------+---------+
| Column Name | Type |
+------------------+---------+
| employee_id | int |
| name | varchar |
| experience_years | int |
+------------------+---------+
employee_id is the primary key of this table.

Write an SQL query that reports all the projects that have the most employees.

The query result format is in the following example:

Project table:
+-------------+-------------+
| project_id | employee_id |
+-------------+-------------+
| 1 | 1 |
| 1 | 2 |
| 1 | 3 |
| 2 | 1 |
| 2 | 4 |
+-------------+-------------+
Employee table:
+-------------+--------+------------------+
| employee_id | name | experience_years |
+-------------+--------+------------------+
| 1 | Khaled | 3 |
| 2 | Ali | 2 |
| 3 | John | 1 |
| 4 | Doe | 2 |
+-------------+--------+------------------+
Result table:
+-------------+
| project_id |
+-------------+
| 1 |
+-------------+
The first project has 3 employees while the second one has 2.
sql
SELECT project_id
FROM Project
GROUP BY project_id
HAVING COUNT(employee_id) = (SELECT COUNT(employee_id)
FROM Project
GROUP BY project_id
ORDER BY COUNT(employee_id) DESC
LIMIT 1)

1077. Project Employees III | Medium | πŸ”’ LeetCode

Table: Project

+-------------+---------+
| Column Name | Type |
+-------------+---------+
| project_id | int |
| employee_id | int |
+-------------+---------+
(project_id, employee_id) is the primary key of this table.
employee_id is a foreign key to Employee table.

Table: Employee

+------------------+---------+
| Column Name | Type |
+------------------+---------+
| employee_id | int |
| name | varchar |
| experience_years | int |
+------------------+---------+
employee_id is the primary key of this table.

Write an SQL query that reports the most experienced employees in each project. In case of a tie, report all employees with the maximum number of experience years.

The query result format is in the following example:

Project table:
+-------------+-------------+
| project_id | employee_id |
+-------------+-------------+
| 1 | 1 |
| 1 | 2 |
| 1 | 3 |
| 2 | 1 |
| 2 | 4 |
+-------------+-------------+
Employee table:
+-------------+--------+------------------+
| employee_id | name | experience_years |
+-------------+--------+------------------+
| 1 | Khaled | 3 |
| 2 | Ali | 2 |
| 3 | John | 3 |
| 4 | Doe | 2 |
+-------------+--------+------------------+
Result table:
+-------------+---------------+
| project_id | employee_id |
+-------------+---------------+
| 1 | 1 |
| 1 | 3 |
| 2 | 1 |
+-------------+---------------+
Both employees with id 1 and 3 have the most experience among the employees of the first project. For the second project, the employee with id 1 has the most experience.

Solution

sql
SELECT
p.project_id,
e.employee_id
FROM
Project p LEFT JOIN Employee e ON
p.employee_id = e.employee_id
WHERE (p.project_id,
e.experience_years) IN (SELECT
p.project_id,
MAX(e.experience_years)
FROM
Project p JOIN Employee e ON
p.employee_id = e.employee_id
GROUP BY
p.project_id)

1082. Sales Analysis I | Easy | πŸ”’ LeetCode

Table: Product

+--------------+---------+
| Column Name | Type |
+--------------+---------+
| product_id | int |
| product_name | varchar |
| unit_price | int |
+--------------+---------+
product_id is the primary key of this table.

Table: Sales

+-------------+---------+
| Column Name | Type |
+-------------+---------+
| seller_id | int |
| product_id | int |
| buyer_id | int |
| sale_date | date |
| quantity | int |
| price | int |
+------ ------+---------+
This table has no primary key, it can have repeated rows.
product_id is a foreign key to Product table.

Write an SQL query that reports the best seller by total sales price, If there is a tie, report them all.

The query result format is in the following example:

Product table:
+------------+--------------+------------+
| product_id | product_name | unit_price |
+------------+--------------+------------+
| 1 | S8 | 1000 |
| 2 | G4 | 800 |
| 3 | iPhone | 1400 |
+------------+--------------+------------+
Sales table:
+-----------+------------+----------+------------+----------+-------+
| seller_id | product_id | buyer_id | sale_date | quantity | price |
+-----------+------------+----------+------------+----------+-------+
| 1 | 1 | 1 | 2019-01-21 | 2 | 2000 |
| 1 | 2 | 2 | 2019-02-17 | 1 | 800 |
| 2 | 2 | 3 | 2019-06-02 | 1 | 800 |
| 3 | 3 | 4 | 2019-05-13 | 2 | 2800 |
+-----------+------------+----------+------------+----------+-------+
Result table:
+-------------+
| seller_id |
+-------------+
| 1 |
| 3 |
+-------------+
Both sellers with id 1 and 3 sold products with the most total price of 2800.

Solution

sql
SELECT seller_id
FROM Sales
GROUP BY seller_id
HAVING SUM(price) = (SELECT SUM(price)
FROM Sales
GROUP BY seller_id
ORDER BY SUM(price) DESC
LIMIT 1)

1083. Sales Analysis II | Easy | πŸ”’ LeetCode

Table: Product

+--------------+---------+
| Column Name | Type |
+--------------+---------+
| product_id | int |
| product_name | varchar |
| unit_price | int |
+--------------+---------+
product_id is the primary key of this table.

Table: Sales

+-------------+---------+
| Column Name | Type |
+-------------+---------+
| seller_id | int |
| product_id | int |
| buyer_id | int |
| sale_date | date |
| quantity | int |
| price | int |
+------ ------+---------+
This table has no primary key, it can have repeated rows.
product_id is a foreign key to Product table.

Write an SQL query that reports the buyers who have bought S8 but not iPhone. Note that S8 and iPhone are products present in the Product table.

The query result format is in the following example:

Product table:
+------------+--------------+------------+
| product_id | product_name | unit_price |
+------------+--------------+------------+
| 1 | S8 | 1000 |
| 2 | G4 | 800 |
| 3 | iPhone | 1400 |
+------------+--------------+------------+
Sales table:
+-----------+------------+----------+------------+----------+-------+
| seller_id | product_id | buyer_id | sale_date | quantity | price |
+-----------+------------+----------+------------+----------+-------+
| 1 | 1 | 1 | 2019-01-21 | 2 | 2000 |
| 1 | 2 | 2 | 2019-02-17 | 1 | 800 |
| 2 | 1 | 3 | 2019-06-02 | 1 | 800 |
| 3 | 3 | 3 | 2019-05-13 | 2 | 2800 |
+-----------+------------+----------+------------+----------+-------+
Result table:
+-------------+
| buyer_id |
+-------------+
| 1 |
+-------------+
The buyer with id 1 bought an S8 but didn't buy an iPhone. The buyer with id 3 bought both.

Solution

sql
SELECT DISTINCT s.buyer_id
FROM Sales s LEFT JOIN Product p ON
s.product_id = p.product_id
WHERE p.product_name = 'S8' AND
s.buyer_id NOT IN (SELECT s.buyer_id
FROM Sales s LEFT JOIN Product p ON
s.product_id = p.product_id
WHERE p.product_name = 'iPhone')

1084. Sales Analysis III | Easy | πŸ”’ LeetCode

Reports the products that were only sold in spring 2019. That is, between 2019-01-01 and 2019-03-31 inclusive. Select the product that were only sold in spring 2019.

Product table:
+------------+--------------+------------+
| product_id | product_name | unit_price |
+------------+--------------+------------+
| 1 | S8 | 1000 |
| 2 | G4 | 800 |
| 3 | iPhone | 1400 |
+------------+--------------+------------+
Sales table:
+-----------+------------+----------+------------+----------+-------+
| seller_id | product_id | buyer_id | sale_date | quantity | price |
+-----------+------------+----------+------------+----------+-------+
| 1 | 1 | 1 | 2019-01-21 | 2 | 2000 |
| 1 | 2 | 2 | 2019-02-17 | 1 | 800 |
| 2 | 2 | 3 | 2019-06-02 | 1 | 800 |
| 3 | 3 | 4 | 2019-05-13 | 2 | 2800 |
+-----------+------------+----------+------------+----------+-------+
Result table:
+-------------+--------------+
| product_id | product_name |
+-------------+--------------+
| 1 | S8 |
+-------------+--------------+
The product with id 1 was only sold in spring 2019 while the other two were sold after.

Solution

sql
(SELECT DISTINCT s.product_id, p.product_name
FROM Sales s LEFT JOIN Product p ON
s.product_id = p.product_id
WHERE s.sale_date >= '2019-01-01' AND
s.sale_date <= '2019-03-31')
EXCEPT -- MINUS if Oracle
(SELECT DISTINCT s.product_id, p.product_name
FROM Sales s LEFT JOIN Product p ON
s.product_id = p.product_id
WHERE s.sale_date < '2019-01-01' OR
s.sale_date > '2019-03-31')

1097. Game Play Analysis V | Hard | πŸ”’ LeetCode

We define the install date of a player to be the first login day of that player. We also define day 1 retention of some date X to be the number of players whose install date is X and they logged back in on the day right after X , divided by the number of players whose install date is X, rounded to 2 decimal places. Write an SQL query that reports for each install date, the number of players that installed the game on that day and the day 1 retention. The query result format is in the following example:

Activity table:
+-----------+-----------+------------+--------------+
| player_id | device_id | event_date | games_played |
+-----------+-----------+------------+--------------+
| 1 | 2 | 2016-03-01 | 5 |
| 1 | 2 | 2016-03-02 | 6 |
| 2 | 3 | 2017-06-25 | 1 |
| 3 | 1 | 2016-03-01 | 0 |
| 3 | 4 | 2016-07-03 | 5 |
+-----------+-----------+------------+--------------+
Result table:
+------------+----------+----------------+
| install_dt | installs | Day1_retention |
+------------+----------+----------------+
| 2016-03-01 | 2 | 0.50 |
| 2017-06-25 | 1 | 0.00 |
+------------+----------+----------------+
Player 1 and 3 installed the game on 2016-03-01 but only player 1 logged back in on 2016-03-02 so the day 1 retention of 2016-03-01 is 1/2 = 0.50
Player 2 installed the game on 2017 -06-25 but didn't log back in on 2017-06-26 so the day 1 retention of 2017-06-25 is 0/1 = 0.00

Solution

sql
SELECT
install_dt,
COUNT(player_id) installs,
ROUND(COUNT(retention)/COUNT(player_id),2) Day1_retention --the number of record on the next day / the total number of id on the day = retention rate
FROM