HackerRank 공부 2

https://www.hackerrank.com/challenges/what-type-of-triangle/problem

Type of Triangle | HackerRank

Sample Input

Sample Output

Isosceles
Equilateral
Scalene
Not A Triangle

 

문제 핵심

 

1. case when 활용해서 컬럼 생성하기

2. case when 순서를 활용해서 집합 관계로 컬럼 생성하기

 

---

내 풀이

 

SELECT
  CASE
    WHEN A + B <= C OR B + C <= A OR A + C <= B THEN 'Not A Triangle'
    WHEN A = B AND B = C THEN 'Equilateral'
    WHEN A = B OR B = C OR A = C THEN 'Isosceles'
    ELSE 'Scalene'
  END AS Type
FROM TRIANGLES;

알게 된 점

---

CASE WHEN에서 VALUE끼리 포함관계의 집합이 있다면 가장 작은 집합부터 부여해 나가자

 

 

---

https://www.hackerrank.com/challenges/occupations/problem

Occupations | HackerRank

 

문제 핵심

Sample Output

Jenny    Ashley     Meera  Jane
Samantha Christeen  Priya  Julia
NULL     Ketty      NULL   Maria

Explanation

The first column is an alphabetically ordered list of Doctor names.
The second column is an alphabetically ordered list of Professor names.
The third column is an alphabetically ordered list of Singer names.
The fourth column is an alphabetically ordered list of Actor names.
The empty cell data for columns with less than the maximum number of names per occupation (in this case, the Professor and Actor columns) are filled with NULL values.

 

내 풀이

---

WITH TEMP
AS
(SELECT Name, Occupation, ROW_NUMBER() OVER(PARTITION BY Occupation ORDER BY Name) AS N
 FROM OCCUPATIONS)
 
 SELECT T1.NAME, T2.NAME, T3.NAME, T4.NAME
 FROM (SELECT NAME, N
       FROM TEMP
        WHERE Occupation = 'Doctor'
        ) AS T1
RIGHT OUTER JOIN
(SELECT NAME, N
       FROM TEMP
        WHERE Occupation = 'Professor'
        ) AS T2
on T1.N = T2.N
LEFT OUTER JOIN
(SELECT NAME, N
       FROM TEMP
        WHERE Occupation = 'Singer'
        ) AS T3
on T2.N = T3.N
LEFT OUTER JOIN
(SELECT NAME, N
       FROM TEMP
        WHERE Occupation = 'Actor'
        ) AS T4
on T2.N = T4.N
;

 

WITH절로 ROW NUMBER 만든 테이블 생성해 두고, ROW NUMBER 기준으로 테이블 JOIN 하면서 NULL 생성함

OUTER JOIN이 없어서 가장 긴 컬럼 직접 확인해서 JOIN 걸었음

 

다른 사람 풀이

---

SELECT min(CASE WHEN Occupation = 'Doctor' THEN Name END) AS 'Doctor',
       max(CASE WHEN Occupation = 'Professor' THEN Name END) AS 'Professor',
       min(CASE WHEN Occupation = 'Singer' THEN Name END) AS 'Singer',
       max(CASE WHEN Occupation = 'Actor' THEN Name END) AS 'Actor'
FROM (
SELECT NAME, OCCUPATION, rank() OVER(PARTITION BY OCCUPATION ORDER BY NAME) AS rank_
FROM OCCUPATIONS
) AS t 
GROUP BY rank_

• NULL을 생성하는 방법으로 GROUP BY도 가능함
• MIN, MAX는 하나의 Name을 가져오기 위한 aggregation 함수
• case when에서 컬럼 = 0 then 다른 컬럼  -> 이 구조 가능함