Leetcode-SQL-

The Leetcode questions and answers

No.1 Select the customer which is not concluded in the order table ( where/ not in)

183. Customers Who Never Order

SELECT Name AS Customers
FROM Customers
WHERE Id NOT IN
(SELECT CustomerId  FROM Orders);

No.2 Select the people who has the highest salary in every department.

184. Department Highest Salary

Tip is joining the two tables and filter the salary with a subquery using MAX() and Group by(). Notice that we should define the salary in a specific department, so we need to select department as well.

SELECT d.Name AS Department,e.Name AS Employee,Salary
FROM Employee e
JOIN Department d
ON e.DepartmentId=d.Id
WHERE (DepartmentId, Salary) IN
(select DepartmentId, max(Salary)
 FROM Employee
 GROUP BY DepartmentId
)      # here we need to select the employee's name, so we cannot use groupby() directly.

No.3 Query the top n things

185. Department Top Three Salaries

When we want to query the top n things, we need to join the same table once, and use 'where n > select count(distinct ) from table where table1. >table2.* ' and in where they ranked the top. This is the key.

SELECT d.Name AS Department, e1.Name AS Employee, Salary
FROM Employee e1
LEFT JOIN Department d
ON e1.DepartmentId=d.Id
WHERE 3 >
(SELECT COUNT(DISTINCT e2.Salary)   # distinct here!
FROM Employee e2
WHERE e2.Salary > e1.Salary
AND e1.DepartmentId=e2.DepartmentId)

No.4 Select the colunmn with more than N things

NO.4 Classes More Than 5 Students

SELECT class 
FROM  
(SELECT class, COUNT(DISTINCT student) AS num
 FROM courses
GROUP BY class) a
WHERE num >=5;

No.5 Delete the replicates which have larger row number/ id

196. Delete Duplicate Emails

Here we use delete to drop out the replicates. And we delte from two same tables so that we can use where to filter the id problem

DELETE p1
FROM Person p1,
    Person p2
WHERE
    p1.Email = p2.Email AND p1.Id > p2.Id
;

No.6 Find out the the id which has rising tempreture compared with its former month

197. Rising Temperature

The idea could be the same with the problems above. So here we can just use the table twice and where clause to realize it.

select w1.Id
from Weather w1, Weather w2
where w1.RecordDate = w2.RecordDate+1 AND w1.Temperature > w2.Temperature;

Or we can use join on the same table to filter the data

SELECT
  weather.id AS 'Id'
FROM
  weather
      JOIN
  weather w ON DATEDIFF(weather.date, w.date) = 1
      AND weather.Temperature > w.Temperature
;

NO.7 Qeury the nth one in the database

176. Second Highest Salary

Here, we can use order by and limit offset to filter the data.
limit means we read how many lines from the top, offset means we skip how many rows to read

SELECT Salary AS SecondHighestSalary
FROM 
(SELECT DISTINCT Salary FROM 
 Employee
 ORDER BY Salary DESC
 limit 1,1) sub ;

SELECT
    (SELECT DISTINCT
            Salary
        FROM
            Employee
        ORDER BY Salary DESC
        LIMIT 1 OFFSET 1) AS SecondHighestSalary
;

NO.8 Find out the duplicate emails

182. Duplicate Emails

here we use Group by to handle this problem, remember use having after the where clause

select Email
from Person
group by Email
having count(Email) > 1;

NO.9 For this kind of ranking question

178. Rank Scores

we just need to compare the score using where to filter the data. And what we should keep in mind is that we need to use distinct for same scores will go into same ranking.

SELECT s.Score,
(SELECT COUNT(DISTINCT Score) FROM Scores WHERE Score >= s.Score) Rank
FROM Scores s
ORDER BY s.Score DESC

NO.10 Exchange the value with adjacent row

626. Exchange Seats

This problem examined mod() and case when. If we want to exchange the adjacent seats, we need to figure out the odd and even number and +1,-1 respectively. Notice we need to know how many records it have, when the last number is odd, we should take the last one out and keep its original id.

SELECT CASE WHEN mod(id,2) !=0 AND id = counts THEN id 
        WHEN mod(id,2) !=0 THEN id+1 
        ELSE id-1 END AS id ,student
FROM seat,(
SELECT COUNT(id) AS counts
    FROM seat) AS seat_count
order by id

HakerRank Problem

No.1 query the city with maximal or minimal length of name.

Here we need to use union to link them. And we use limit 1 to choose the first one. And when it comes to maximal or minimal, we can first think about using order by. Because when we want to use Max or Min, we need to use Group by

(SELECT CITY, LENGTH(CITY) 
FROM STATION 
ORDER BY LENGTH(CITY), CITY LIMIT 1) 
UNION 
(SELECT CITY, LENGTH(CITY) FROM STATION 
ORDER BY LENGTH(CITY) DESC, CITY ASC LIMIT 1);

No.2 Fuzzy or accurate search

here we need to understand the difference between Like and Regexp

Regexp search data having one of the letter in the beginning of the value like ^[wk]

SELECT DISTINCT CITY
FROM STATION 
WHERE LOWER(CITY) REGEXP '^[aeiou]';

Regexp search data having the condition in the end of the value like [ww]$

SELECT DISTINCT CITY
FROM STATION
WHERE LOWER(CITY) REGEXP '[aeiou]$'

Regexp search data having the condition in the both beginning and end of the value like ^[ek].*[12]$

SELECT DISTINCT city FROM station WHERE city RLIKE '^[aeiou].*[aeiou]$'
SELECT DISTINCT City
FROM Station
WHERE City RLIKE '^[^AEIOU].*[^aeiou]$';

Or we can query data like this

select distinct city from station 
where left(city,1) in ('a','e','i','o','u') 
and right(city, 1) in ('a','e','i','o','u')

not start with aeiou

SELECT DISTINCT CITY FROM STATION
WHERE CITY REGEXP '^[^aeiou]'

select DISTINCT city from STATION where city NOT REGEXP '[aeiou]$'

No.3 How to write a function in MYSQL

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

Practice

Practice: 3 Tables
S: s_no, s_name
C: c_no, t_name
SC: s_no, c_no, grade

Q1 : find out the students who didn't choose Li's course

select s_name from S
where s_no not in 
(
select sno from SC where c_no in
(select c_no from C 
where t_name='lining')
)

Q2 : find out the students who have more than 2 courses less than 60

select name
from
(
select s_no 
from SC
where grade <60
group by s_no
having count(s_no) >=2
)a
join 
(
select s_no, name 
from S
)b
on a.s_no=b.s_no

Q3 : find out students who choose course1 and course2

select name 
from S
where s_no in 
(
select * 
from 
(
select s_no 
from SC 
where c_no=1
)a
join
(
select s_no
from SC
where c_no=2
)b
on a.s_no = b.s_no
)

面试

Q1 : 有一张active表，字段有date, id，算出每日活跃，次日留存，7日留存 Tips想要拿到留存，就得知道每天新增用户，以及次日还在活跃的用户，7日还在活跃的用户有哪些，主要用到join和datediff（）

SELECT ymd
	   ,count(distinct id_0) as "active_users"
	   ,count(distinct id_1)/count(distinct id_0) as "1day_renten"
	   ,count(distinct id_2)/count(distinct id_0) as "2day_renten"
FROM
	   (SELECT a.ymd
		       ,a.id AS id_0
		       ,b.id AS id_1
		       ,c.id AS id_2
		FROM active a
		LEFT JOIN active b
		ON datediff(date(b.ymd),date(a.ymd))=1 -- 次日留存
		AND b.id=a.id
		LEFT JOIN active c
		ON datediff(date(c.ymd),date(a.ymd))=2 -- 3日留存
		AND c.id=a.id
		ORDER BY a.ymd)table1
GROUP BY ymd

nicolehhy / Leetcode-SQL-