LeetCode SQL Problems ①

Easy

1179. Reformat Department Table

Table:

+---------------+---------+
| Column Name | Type |
+---------------+---------+
| id | int |
| revenue | int |
| month | varchar |
+---------------+---------+
(id, month) is the primary key of this table.
The table has information about the revenue of each department per month.
The month has values in ["Jan","Feb","Mar","Apr","May","Jun","Jul","Aug","Sep","Oct","Nov","Dec"].

Write an SQL query to reformat the table such that there is a department id column and a revenue column for each month.

The query result format is in the following example:

Department table:
+------+---------+-------+
| id | revenue | month |
+------+---------+-------+
| 1 | 8000 | Jan |
| 2 | 9000 | Jan |
| 3 | 10000 | Feb |
| 1 | 7000 | Feb |
| 1 | 6000 | Mar |
+------+---------+-------+
Result table:
+------+-------------+-------------+-------------+-----+-------------+
| id | Jan_Revenue | Feb_Revenue | Mar_Revenue | ... | Dec_Revenue |
+------+-------------+-------------+-------------+-----+-------------+
| 1 | 8000 | 7000 | 6000 | ... | null |
| 2 | 9000 | null | null | ... | null |
| 3 | null | 10000 | null | ... | null |
+------+-------------+-------------+-------------+-----+-------------+
Note that the result table has 13 columns (1 for the department id + 12 for the months).

Solution

SELECT id,
SUM(CASE WHEN month="Jan" THEN revenue END) AS "Jan_Revenue", SUM(CASE WHEN month="Mar" THEN revenue END) AS "Mar_Revenue", SUM(CASE WHEN month="Apr" THEN revenue END) AS "Apr_Revenue", SUM(CASE WHEN month="Feb" THEN revenue END) AS "Feb_Revenue", SUM(CASE WHEN month="May" THEN revenue END) AS "May_Revenue", SUM(CASE WHEN month="Jun" THEN revenue END) AS "Jun_Revenue", SUM(CASE WHEN month="Jul" THEN revenue END) AS "Jul_Revenue", SUM(CASE WHEN month="Aug" THEN revenue END) AS "Aug_Revenue", SUM(CASE WHEN month="Sep" THEN revenue END) AS "Sep_Revenue", SUM(CASE WHEN month="Oct" THEN revenue END) AS "Oct_Revenue", SUM(CASE WHEN month="Nov" THEN revenue END) AS "Nov_Revenue", SUM(CASE WHEN month="Dec" THEN revenue END) AS "Dec_Revenue"
FROM department
GROUP BY id

183. Customers Who Never Order

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

Table: .

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

Table: .

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

Using the above tables as example, return the following:

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

Solution

SELECT c.name Customers 
FROM customers c
LEFT JOIN orders o ON c.id=o.customerid
WHERE o.customerid IS NULL

197. Rising Temperature

Table:

+---------------+---------+
| 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’ 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

SELECT w2.id 
FROM weather w1
JOIN weather w2 ON w2.recordDate = DATE_ADD(w1.recordDate, INTERVAL 1 DAY)
WHERE w2.temperature > w1.temperature

181. Employees Earning More Than Their Managers

The 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 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

SELECT e.name Employee 
FROM employee e
INNER JOIN employee m ON e.managerID = m.ID
WHERE e.salary > m.salary

627. Swap Salary

Given a table , such as the one below, that has m=male and f=female values. 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.

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

Example:

| id | name | sex | salary |
|----|------|-----|--------|
| 1 | A | m | 2500 |
| 2 | B | f | 1500 |
| 3 | C | m | 5500 |
| 4 | D | f | 500 |

After running your update statement, the above salary table should have the following rows:

| id | name | sex | salary |
|----|------|-----|--------|
| 1 | A | f | 2500 |
| 2 | B | m | 1500 |
| 3 | C | f | 5500 |
| 4 | D | m | 500 |

Solution

UPDATE salary  
SET sex = IF(sex='m','f','m')

196. Delete Duplicate Emails

Write a SQL query to delete all duplicate email entries in a table named , 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 table should have the following rows:

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

Solution

DELETE FROM person WHERE id IN 
(SELECT id
FROM (SELECT *, COUNT(email) OVER (PARTITION BY email) cnt, RANK() OVER (PARTITION BY email ORDER BY id) rnk
FROM person) sub
WHERE cnt > 1 AND rnk > 1)

620. Not Boring Movies

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 :

+---------+-----------+--------------+-----------+
| 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

SELECT * 
FROM cinema
WHERE MOD(id, 2) = 1 AND description <> 'boring'
ORDER BY rating DESC

595. Big Countries

There is a table

+-----------------+------------+------------+--------------+---------------+
| 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

SELECT name, population, area 
FROM world
WHERE area > 3000000 OR population > 25000000

182. Duplicate Emails

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

+----+---------+
| 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 |
+---------+

Solution

SELECT email 
FROM person
GROUP BY email
HAVING COUNT(email) > 1

175. Combine Two Tables

Table:

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

Table:

+-------------+---------+
| 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

SELECT p.firstname, p.lastname, a.city, a.state  
FROM person p
LEFT JOIN address a ON a.personid = p.personid

Keep going!