[MySQL] 리트코드 585. Investments in 2016

문제

Table: Insurance
+-------------+-------+
| Column Name | Type  |
+-------------+-------+
| pid         | int   |
| tiv_2015    | float |
| tiv_2016    | float |
| lat         | float |
| lon         | float |
+-------------+-------+

pid is the primary key (column with unique values) for this table.
Each row of this table contains information about one policy where:
pid is the policyholder's policy ID.
tiv_2015 is the total investment value in 2015 and tiv_2016 is the total investment value in 2016.
lat is the latitude of the policy holder's city. It's guaranteed that lat is not NULL.
lon is the longitude of the policy holder's city. It's guaranteed that lon is not NULL.
 
Write a solution to report the sum of all total investment values in 2016 tiv_2016, for all policyholders who:
  have the same tiv_2015 value as one or more other policyholders, and
  are not located in the same city as any other policyholder (i.e., the (lat, lon) attribute pairs must be unique).
Round tiv_2016 to two decimal places.

 

https://leetcode.com/problems/investments-in-2016/description/

예시

Input: 
Insurance table:
+-----+----------+----------+-----+-----+
| pid | tiv_2015 | tiv_2016 | lat | lon |
+-----+----------+----------+-----+-----+
| 1   | 10       | 5        | 10  | 10  |
| 2   | 20       | 20       | 20  | 20  |
| 3   | 10       | 30       | 20  | 20  |
| 4   | 10       | 40       | 40  | 40  |
+-----+----------+----------+-----+-----+
Output: 
+----------+
| tiv_2016 |
+----------+
| 45.00    |
+----------+

Explanation: 
The first record in the table, like the last record, meets both of the two criteria.
The tiv_2015 value 10 is the same as the third and fourth records, and its location is unique.

The second record does not meet any of the two criteria. Its tiv_2015 is not like any other policyholders and its location is the same as the third record, which makes the third record fail, too.
So, the result is the sum of tiv_2016 of the first and last record, which is 45.

문제 풀이

SELECT
	-- 소수점 둘째 자리까지 출력
    ROUND(SUM(i1.tiv_2016), 2) AS tiv_2016
FROM
    insurance AS i1
WHERE
	-- 조건 1: tiv_2015가 1개 이상
    i1.tiv_2015 IN (
        SELECT
            i2.tiv_2015
        FROM
            insurance AS i2
        GROUP BY
            tiv_2015
        HAVING
            COUNT(tiv_2015) > 1
    )
    -- 조건 2: lat, lon pair가 unique
    AND (i1.lat, i1.lon) IN(
        SELECT
            i3.lat, i3.lon
        FROM
            insurance AS i3
        GROUP BY
            lat, lon
        HAVING
            COUNT(*) = 1
    );

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SQL을 독학하시는 분들에게 도움이 되길 바라며,

혹 더 좋은 방법이 있거나 오류가 있다면 편하게 말씀 부탁드립니다.