Tag: Business Intelligence Tools

Information is the oil of the 21st century, and analytics is the combustion engine.

Peter Sondergaard (Former EVP, Research & Advisory – Gartner)

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Question Source: LeetCode

Solution Language: MySQL

This Q&A series will cover data questions from LeetCode and present my solutions to them. Please feel free to comment with your suggestions if you feel that these problems may be solved in a more optimized manner.

Question (LeetCode Question #1412, Level: Hard)

Table: Student

+---------------------+---------+
| Column Name         | Type    |
+---------------------+---------+
| student_id          | int     |
| student_name        | varchar |
+---------------------+---------+
student_id is the primary key for this table.
student_name is the name of the student.

Table: Exam

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| exam_id       | int     |
| student_id    | int     |
| score         | int     |
+---------------+---------+
(exam_id, student_id) is the primary key for this table.
Student with student_id got score points in exam with id exam_id.

A “quiet” student is the one who took at least one exam and didn’t score neither the high score nor the low score.

Write an SQL query to report the students (student_id, student_name) being “quiet” in ALL exams.

Don’t return the student who has never taken any exam. Return the result table ordered by student_id.

The query result format is in the following example.

Student table:
+-------------+---------------+
| student_id  | student_name  |
+-------------+---------------+
| 1           | Daniel        |
| 2           | Jade          |
| 3           | Stella        |
| 4           | Jonathan      |
| 5           | Will          |
+-------------+---------------+

Exam table:
+------------+--------------+-----------+
| exam_id    | student_id   | score     |
+------------+--------------+-----------+
| 10         |     1        |    70     |
| 10         |     2        |    80     |
| 10         |     3        |    90     |
| 20         |     1        |    80     |
| 30         |     1        |    70     |
| 30         |     3        |    80     |
| 30         |     4        |    90     |
| 40         |     1        |    60     |
| 40         |     2        |    70     |
| 40         |     4        |    80     |
+------------+--------------+-----------+

Result table:
+-------------+---------------+
| student_id  | student_name  |
+-------------+---------------+
| 2           | Jade          |
+-------------+---------------+

For exam 1: Student 1 and 3 hold the lowest and high score respectively.
For exam 2: Student 1 hold both highest and lowest score.
For exam 3 and 4: Studnet 1 and 4 hold the lowest and high score respectively.
Student 2 and 5 have never got the highest or lowest in any of the exam.
Since student 5 is not taking any exam, he is excluded from the result.
So, we only return the information of Student 2.

Solution

WITH cte AS 
(SELECT student_id, 
             score, 
             exam_id,
             (CASE WHEN score < MAX(score) OVER (PARTITION BY exam_id)
             AND score > MIN(score) OVER (PARTITION BY exam_id)
             THEN 'middle'
             ELSE 'highlow'
             END) AS category
FROM Exam
ORDER BY student_id),

cte1 AS (SELECT student_id
         FROM cte
         GROUP BY student_id
         HAVING SUM(CASE WHEN category = 'highlow'
                    THEN 1 ELSE 0
                    END) = 0
)

SELECT cte1.student_id, s.student_name
FROM cte1 JOIN Student s
ON cte1.student_id = s.student_id
ORDER BY cte1.student_id

Alternate Approaches…

WITH cte AS 
     (SELECT student_id, score, exam_id,
            max(score) OVER (PARTITION BY exam_id) AS maxscore,
            min(score) OVER (PARTITION BY exam_id) AS minscore
      FROM Exam),
      cte1 AS 
            (SELECT student_id
                  FROM cte
                  WHERE score = maxscore OR score = minscore
             )
SELECT DISTINCT Exam.student_id, Student.student_name
      FROM Exam JOIN Student
      ON Exam.student_id = Student.student_id
WHERE Exam.student_id NOT IN (SELECT student_id FROM cte1)
      ORDER BY Exam.student_id

WITH cte AS (SELECT student_id,
             rank() OVER (PARTITION BY exam_id ORDER BY score DESC) 
             AS gethighest,
             rank() OVER (PARTITION BY exam_id ORDER BY score ASC) 
             AS getlowest
             FROM Exam),
     cte1 AS (SELECT DISTINCT student_id,
              SUM(CASE WHEN gethighest = 1 
                            OR getlowest = 1 
                       THEN 1 
                       ELSE 0 END) 
                  OVER (PARTITION BY student_id ORDER BY student_id)
                  AS numofhighlow
              FROM cte)
SELECT cte1.student_id, student_name
      FROM cte1 JOIN Student
      ON cte1.student_id = Student.student_id
      WHERE cte1.numofhighlow = 0

If somebody tortures the data enough (open or not), it will confess anything.

Paolo Magrassi, (Former vice president, research director, Gartner)

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Question Source: LeetCode

Solution Language: MySQL

This Q&A series will cover data questions from LeetCode and present my solutions to them. Please feel free to comment with your suggestions if you feel that these problems may be solved in a more optimized manner.

Question (LeetCode Question #262, Level: Hard)

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 (‘completed’, ‘cancelled_by_driver’, ‘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 (‘client’, ‘driver’, ‘partner’).
Banned is an ENUM type of (‘Yes’, ‘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

Approach With Joins:

WITH temp AS
(SELECT DISTINCT t.Request_at AS Day,
COUNT(CASE WHEN Status <> ‘completed’ THEN Id ELSE null END) OVER
(PARTITION BY t.Request_at) AS canceled,
COUNT(Id) OVER(PARTITION BY t.Request_at) AS total
FROM Trips t JOIN Users uc JOIN Users ud
ON t.Client_Id = uc.Users_Id
AND t.Driver_Id = ud.Users_Id
WHERE uc.Banned = ‘No’ AND ud.Banned = ‘No’
AND t.Request_at BETWEEN CAST(‘2013-10-01’ AS Date)
AND CAST(‘2013-10-03’ AS Date))
SELECT Day,
CAST(canceled/total AS DECIMAL(65,2)) AS ‘Cancellation Rate’
FROM temp

Alternate Approach (Without Joins)

SELECT Request_at AS Day,
CAST(COUNT(IF(Status != ‘completed’, true, null)) / COUNT(Id) AS DECIMAL(65,2))
AS ‘Cancellation Rate’
FROM Trips
WHERE Request_at BETWEEN ‘2013-10-01’ AND ‘2013-10-03’
AND Client_id IN (SELECT Users_Id FROM Users WHERE Banned = ‘No’)
AND Driver_Id IN (SELECT Users_Id FROM Users WHERE Banned = ‘No’)
GROUP BY Request_at;