StarMirror/CS-Notes

Fork 0

CyC2018 edbe84cef7 auto commit

2020-05-27 01:39:09 +08:00

22 KiB

Raw Blame History

595. Big Countries
627. Swap Salary
620. Not Boring Movies
596. Classes More Than 5 Students
182. Duplicate Emails
196. Delete Duplicate Emails
175. Combine Two Tables
181. Employees Earning More Than Their Managers
183. Customers Who Never Order
184. Department Highest Salary
176. Second Highest Salary
177. Nth Highest Salary
178. Rank Scores
180. Consecutive Numbers
626. Exchange Seats

595. Big Countries

https://leetcode.com/problems/big-countries/description/

Description

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

查找面积超过 3,000,000 或者人口数超过 25,000,000 的国家。

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

Solution

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

SQL Schema

SQL Schema 用于在本地环境下创建表结构并导入数据，从而方便在本地环境调试。

DROP TABLE
IF
    EXISTS World;
CREATE TABLE World ( NAME VARCHAR ( 255 ), continent VARCHAR ( 255 ), area INT, population INT, gdp INT );
INSERT INTO World ( NAME, continent, area, population, gdp )
VALUES
    ( 'Afghanistan', 'Asia', '652230', '25500100', '203430000' ),
    ( 'Albania', 'Europe', '28748', '2831741', '129600000' ),
    ( 'Algeria', 'Africa', '2381741', '37100000', '1886810000' ),
    ( 'Andorra', 'Europe', '468', '78115', '37120000' ),
    ( 'Angola', 'Africa', '1246700', '20609294', '1009900000' );

627. Swap Salary

https://leetcode.com/problems/swap-salary/description/

Description

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

只用一个 SQL 查询，将 sex 字段反转。

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

Solution

两个相等的数异或的结果为 0，而 0 与任何一个数异或的结果为这个数。

sex 字段只有两个取值：'f' 和 'm'，并且有以下规律：

'f' ^ ('m' ^ 'f') = 'm' ^ ('f' ^ 'f') = 'm'
'm' ^ ('m' ^ 'f') = 'f' ^ ('m' ^ 'm') = 'f'

因此将 sex 字段和 'm' ^ 'f' 进行异或操作，最后就能反转 sex 字段。

UPDATE salary
SET sex = CHAR ( ASCII(sex) ^ ASCII( 'm' ) ^ ASCII( 'f' ) );

SQL Schema

DROP TABLE
IF
    EXISTS salary;
CREATE TABLE salary ( id INT, NAME VARCHAR ( 100 ), sex CHAR ( 1 ), salary INT );
INSERT INTO salary ( id, NAME, sex, salary )
VALUES
    ( '1', 'A', 'm', '2500' ),
    ( '2', 'B', 'f', '1500' ),
    ( '3', 'C', 'm', '5500' ),
    ( '4', 'D', 'f', '500' );

620. Not Boring Movies

https://leetcode.com/problems/not-boring-movies/description/

Description

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

查找 id 为奇数，并且 description 不是 boring 的电影，按 rating 降序。

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

Solution

SELECT
    *
FROM
    cinema
WHERE
    id % 2 = 1
    AND description != 'boring'
ORDER BY
    rating DESC;

SQL Schema

DROP TABLE
IF
    EXISTS cinema;
CREATE TABLE cinema ( id INT, movie VARCHAR ( 255 ), description VARCHAR ( 255 ), rating FLOAT ( 2, 1 ) );
INSERT INTO cinema ( id, movie, description, rating )
VALUES
    ( 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 );

596. Classes More Than 5 Students

https://leetcode.com/problems/classes-more-than-5-students/description/

Description

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

查找有五名及以上 student 的 class。

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

Solution

对 class 列进行分组之后，再使用 count 汇总函数统计每个分组的记录个数，之后使用 HAVING 进行筛选。HAVING 针对分组进行筛选，而 WHERE 针对每个记录（行）进行筛选。

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

SQL Schema

DROP TABLE
IF
    EXISTS courses;
CREATE TABLE courses ( student VARCHAR ( 255 ), class VARCHAR ( 255 ) );
INSERT INTO courses ( student, class )
VALUES
    ( 'A', 'Math' ),
    ( 'B', 'English' ),
    ( 'C', 'Math' ),
    ( 'D', 'Biology' ),
    ( 'E', 'Math' ),
    ( 'F', 'Computer' ),
    ( 'G', 'Math' ),
    ( 'H', 'Math' ),
    ( 'I', 'Math' );

182. Duplicate Emails

https://leetcode.com/problems/duplicate-emails/description/

Description

邮件地址表：

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

查找重复的邮件地址：

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

Solution

对 Email 进行分组，如果并使用 COUNT 进行计数统计，结果大于等于 2 的表示 Email 重复。

SELECT
    Email
FROM
    Person
GROUP BY
    Email
HAVING
    COUNT( * ) >= 2;

SQL Schema

DROP TABLE
IF
    EXISTS Person;
CREATE TABLE Person ( Id INT, Email VARCHAR ( 255 ) );
INSERT INTO Person ( Id, Email )
VALUES
    ( 1, 'a@b.com' ),
    ( 2, 'c@d.com' ),
    ( 3, 'a@b.com' );

196. Delete Duplicate Emails

https://leetcode.com/problems/delete-duplicate-emails/description/

Description

邮件地址表：

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

删除重复的邮件地址：

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

Solution

只保留相同 Email 中 Id 最小的那一个，然后删除其它的。

连接查询：

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

子查询：

DELETE
FROM
    Person
WHERE
    id NOT IN (
        SELECT id 
        FROM ( 
            SELECT min( id ) AS id 
            FROM Person
            GROUP BY email
        ) AS m
    );

应该注意的是上述解法额外嵌套了一个 SELECT 语句，如果不这么做，会出现错误：You can't specify target table 'Person' for update in FROM clause。以下演示了这种错误解法。

DELETE
FROM
    Person
WHERE
    id NOT IN ( 
        SELECT min( id ) AS id 
        FROM Person 
        GROUP BY email 
    );

参考：pMySQL Error 1093 - Can't specify target table for update in FROM clause

SQL Schema

与 182 相同。

175. Combine Two Tables

https://leetcode.com/problems/combine-two-tables/description/

Description

Person 表：

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

Address 表：

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

查找 FirstName, LastName, City, State 数据，而不管一个用户有没有填地址信息。

Solution

涉及到 Person 和 Address 两个表，在对这两个表执行连接操作时，因为要保留 Person 表中的信息，即使在 Address 表中没有关联的信息也要保留。此时可以用左外连接，将 Person 表放在 LEFT JOIN 的左边。

SELECT
    FirstName,
    LastName,
    City,
    State
FROM
    Person P
    LEFT JOIN Address A
    ON P.PersonId = A.PersonId;

SQL Schema

DROP TABLE
IF
    EXISTS Person;
CREATE TABLE Person ( PersonId INT, FirstName VARCHAR ( 255 ), LastName VARCHAR ( 255 ) );
DROP TABLE
IF
    EXISTS Address;
CREATE TABLE Address ( AddressId INT, PersonId INT, City VARCHAR ( 255 ), State VARCHAR ( 255 ) );
INSERT INTO Person ( PersonId, LastName, FirstName )
VALUES
    ( 1, 'Wang', 'Allen' );
INSERT INTO Address ( AddressId, PersonId, City, State )
VALUES
    ( 1, 2, 'New York City', 'New York' );

181. Employees Earning More Than Their Managers

https://leetcode.com/problems/employees-earning-more-than-their-managers/description/

Description

Employee 表：

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

查找薪资大于其经理薪资的员工信息。

Solution

SELECT
    E1.NAME AS Employee
FROM
    Employee E1
    INNER JOIN Employee E2
    ON E1.ManagerId = E2.Id
    AND E1.Salary > E2.Salary;

SQL Schema

DROP TABLE
IF
    EXISTS Employee;
CREATE TABLE Employee ( Id INT, NAME VARCHAR ( 255 ), Salary INT, ManagerId INT );
INSERT INTO Employee ( Id, NAME, Salary, ManagerId )
VALUES
    ( 1, 'Joe', 70000, 3 ),
    ( 2, 'Henry', 80000, 4 ),
    ( 3, 'Sam', 60000, NULL ),
    ( 4, 'Max', 90000, NULL );

183. Customers Who Never Order

https://leetcode.com/problems/customers-who-never-order/description/

Description

Customers 表：

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

Orders 表：

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

查找没有订单的顾客信息：

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

Solution

左外链接

SELECT
    C.Name AS Customers
FROM
    Customers C
    LEFT JOIN Orders O
    ON C.Id = O.CustomerId
WHERE
    O.CustomerId IS NULL;

子查询

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

SQL Schema

DROP TABLE
IF
    EXISTS Customers;
CREATE TABLE Customers ( Id INT, NAME VARCHAR ( 255 ) );
DROP TABLE
IF
    EXISTS Orders;
CREATE TABLE Orders ( Id INT, CustomerId INT );
INSERT INTO Customers ( Id, NAME )
VALUES
    ( 1, 'Joe' ),
    ( 2, 'Henry' ),
    ( 3, 'Sam' ),
    ( 4, 'Max' );
INSERT INTO Orders ( Id, CustomerId )
VALUES
    ( 1, 3 ),
    ( 2, 1 );

184. Department Highest Salary

https://leetcode.com/problems/department-highest-salary/description/

Description

Employee 表：

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

Department 表：

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

查找一个 Department 中收入最高者的信息：

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

Solution

创建一个临时表，包含了部门员工的最大薪资。可以对部门进行分组，然后使用 MAX() 汇总函数取得最大薪资。

之后使用连接找到一个部门中薪资等于临时表中最大薪资的员工。

SELECT
    D.NAME Department,
    E.NAME Employee,
    E.Salary
FROM
    Employee E,
    Department D,
    ( SELECT DepartmentId, MAX( Salary ) Salary 
     FROM Employee 
     GROUP BY DepartmentId ) M
WHERE
    E.DepartmentId = D.Id
    AND E.DepartmentId = M.DepartmentId
    AND E.Salary = M.Salary;

SQL Schema

DROP TABLE IF EXISTS Employee;
CREATE TABLE Employee ( Id INT, NAME VARCHAR ( 255 ), Salary INT, DepartmentId INT );
DROP TABLE IF EXISTS Department;
CREATE TABLE Department ( Id INT, NAME VARCHAR ( 255 ) );
INSERT INTO Employee ( Id, NAME, Salary, DepartmentId )
VALUES
    ( 1, 'Joe', 70000, 1 ),
    ( 2, 'Henry', 80000, 2 ),
    ( 3, 'Sam', 60000, 2 ),
    ( 4, 'Max', 90000, 1 );
INSERT INTO Department ( Id, NAME )
VALUES
    ( 1, 'IT' ),
    ( 2, 'Sales' );

176. Second Highest Salary

https://leetcode.com/problems/second-highest-salary/description/

Description

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

查找工资第二高的员工。

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

没有找到返回 null 而不是不返回数据。

Solution

为了在没有查找到数据时返回 null，需要在查询结果外面再套一层 SELECT。

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

SQL Schema

DROP TABLE
IF
    EXISTS Employee;
CREATE TABLE Employee ( Id INT, Salary INT );
INSERT INTO Employee ( Id, Salary )
VALUES
    ( 1, 100 ),
    ( 2, 200 ),
    ( 3, 300 );

177. Nth Highest Salary

Description

查找工资第 N 高的员工。

Solution

CREATE FUNCTION getNthHighestSalary ( N INT ) RETURNS INT BEGIN

SET N = N - 1;
RETURN ( 
    SELECT ( 
        SELECT DISTINCT Salary 
        FROM Employee 
        ORDER BY Salary DESC 
        LIMIT N, 1 
    ) 
);

END

SQL Schema

同 176。

178. Rank Scores

https://leetcode.com/problems/rank-scores/description/

Description

得分表：

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

将得分排序，并统计排名。

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

Solution

要统计某个 score 的排名，只要统计大于等于该 score 的 score 数量。

Id	score	大于等于该 score 的 score 数量	排名
1	4.1	3	3
2	4.2	2	2
3	4.3	1	1

使用连接操作找到某个 score 对应的大于等于其值的记录：

SELECT
	*
FROM
    Scores S1
    INNER JOIN Scores S2
    ON S1.score <= S2.score
ORDER BY
    S1.score DESC, S1.Id;

S1.Id	S1.score	S2.Id	S2.score
3	4.3	3	4.3
2	4.2	2	4.2
2	4.2	3	4.3
1	4.1	1	4.1
1	4.1	2	4.2
1	4.1	3	4.3

可以看到每个 S1.score 都有对应好几条记录，我们再进行分组，并统计每个分组的数量作为 'Rank'

SELECT
    S1.score 'Score',
    COUNT(*) 'Rank'
FROM
    Scores S1
    INNER JOIN Scores S2
    ON S1.score <= S2.score
GROUP BY
    S1.id, S1.score
ORDER BY
    S1.score DESC, S1.Id;

score	Rank
4.3	1
4.2	2
4.1	3

上面的解法看似没问题，但是对于以下数据，它却得到了错误的结果：

Id	score
1	4.1
2	4.2
3	4.2

score	Rank
4.2	2
4.2	2
4.1	3

而我们希望的结果为：

score	Rank
4.2	1
4.2	1
4.1	2

连接情况如下：

S1.Id	S1.score	S2.Id	S2.score
2	4.2	3	4.2
2	4.2	2	4.2
3	4.2	3	4.2
3	4.2	2	4.1
1	4.1	3	4.2
1	4.1	2	4.2
1	4.1	1	4.1

我们想要的结果是，把分数相同的放在同一个排名，并且相同分数只占一个位置，例如上面的分数，Id=2 和 Id=3 的记录都有相同的分数，并且最高，他们并列第一。而 Id=1 的记录应该排第二名，而不是第三名。所以在进行 COUNT 计数统计时，我们需要使用 COUNT( DISTINCT S2.score ) 从而只统计一次相同的分数。

SELECT
    S1.score 'Score',
    COUNT( DISTINCT S2.score ) 'Rank'
FROM
    Scores S1
    INNER JOIN Scores S2
    ON S1.score <= S2.score
GROUP BY
    S1.id, S1.score
ORDER BY
    S1.score DESC;

SQL Schema

DROP TABLE
IF
    EXISTS Scores;
CREATE TABLE Scores ( Id INT, Score DECIMAL ( 3, 2 ) );
INSERT INTO Scores ( Id, Score )
VALUES
    ( 1, 4.1 ),
    ( 2, 4.1 ),
    ( 3, 4.2 ),
    ( 4, 4.2 ),
    ( 5, 4.3 ),
    ( 6, 4.3 );

180. Consecutive Numbers

https://leetcode.com/problems/consecutive-numbers/description/

Description

数字表：

+----+-----+
| Id | Num |
+----+-----+
| 1  |  1  |
| 2  |  1  |
| 3  |  1  |
| 4  |  2  |
| 5  |  1  |
| 6  |  2  |
| 7  |  2  |
+----+-----+

查找连续出现三次的数字。

+-----------------+
| ConsecutiveNums |
+-----------------+
| 1               |
+-----------------+

Solution

SELECT
    DISTINCT L1.num ConsecutiveNums
FROM
    Logs L1,
    Logs L2,
    Logs L3
WHERE L1.id = l2.id - 1
    AND L2.id = L3.id - 1
    AND L1.num = L2.num
    AND l2.num = l3.num;

SQL Schema

DROP TABLE
IF
    EXISTS LOGS;
CREATE TABLE LOGS ( Id INT, Num INT );
INSERT INTO LOGS ( Id, Num )
VALUES
    ( 1, 1 ),
    ( 2, 1 ),
    ( 3, 1 ),
    ( 4, 2 ),
    ( 5, 1 ),
    ( 6, 2 ),
    ( 7, 2 );

626. Exchange Seats

https://leetcode.com/problems/exchange-seats/description/

Description

seat 表存储着座位对应的学生。

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

要求交换相邻座位的两个学生，如果最后一个座位是奇数，那么不交换这个座位上的学生。

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

Solution

使用多个 union。

# 处理偶数 id，让 id 减 1
# 例如 2,4,6,... 变成 1,3,5,...
SELECT
    s1.id - 1 AS id,
    s1.student
FROM
    seat s1
WHERE
    s1.id MOD 2 = 0 UNION
# 处理奇数 id，让 id 加 1。但是如果最大的 id 为奇数，则不做处理
# 例如 1,3,5,... 变成 2,4,6,...
SELECT
    s2.id + 1 AS id,
    s2.student
FROM
    seat s2
WHERE
    s2.id MOD 2 = 1
    AND s2.id != ( SELECT max( s3.id ) FROM seat s3 ) UNION
# 如果最大的 id 为奇数，单独取出这个数
SELECT
    s4.id AS id,
    s4.student
FROM
    seat s4
WHERE
    s4.id MOD 2 = 1
    AND s4.id = ( SELECT max( s5.id ) FROM seat s5 )
ORDER BY
    id;

SQL Schema

DROP TABLE
IF
    EXISTS seat;
CREATE TABLE seat ( id INT, student VARCHAR ( 255 ) );
INSERT INTO seat ( id, student )
VALUES
    ( '1', 'Abbot' ),
    ( '2', 'Doris' ),
    ( '3', 'Emerson' ),
    ( '4', 'Green' ),
    ( '5', 'Jeames' );

22 KiB Raw Blame History Unescape Escape

595. Big Countries

Description

Solution

SQL Schema

627. Swap Salary

Description

Solution

SQL Schema

620. Not Boring Movies

Description

Solution

SQL Schema

596. Classes More Than 5 Students

Description

Solution

SQL Schema

182. Duplicate Emails

Description

Solution

SQL Schema

196. Delete Duplicate Emails

Description

Solution

SQL Schema

175. Combine Two Tables

Description

Solution

SQL Schema

181. Employees Earning More Than Their Managers

Description

Solution

SQL Schema

183. Customers Who Never Order

Description

Solution

SQL Schema

184. Department Highest Salary

Description

Solution

SQL Schema

176. Second Highest Salary

Description

Solution

SQL Schema

177. Nth Highest Salary

Description

Solution

SQL Schema

178. Rank Scores

Description

Solution

SQL Schema

180. Consecutive Numbers

Description

Solution

SQL Schema

626. Exchange Seats

Description

Solution

SQL Schema

22 KiB

Raw Blame History