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1 change: 1 addition & 0 deletions README.md
Original file line number Diff line number Diff line change
Expand Up @@ -83,6 +83,7 @@ Have a good contributing!
- [184. Department Highest Salary](./leetcode/medium/184.%20Department%20Highest%20Salary.sql)
- [550. Game Play Analysis IV](./leetcode/medium/550.%20Game%20Play%20Analysis%20IV.sql)
- [570. Managers with at Least 5 Direct Reports](./leetcode/medium/570.%20Managers%20with%20at%20Least%205%20Direct%20Reports.sql)
- [585. Investments in 2016](./leetcode/medium/585.%20Investments%20in%202016.sql)
- [602. Friend Requests II: Who Has the Most Friends](./leetcode/medium/602.%20Friend%20Requests%20II:%20Who%20Has%20the%20Most%20Friends.sql)
- [626. Exchange Seats](./leetcode/medium/626.%20Exchange%20Seats.sql)
- [1045. Customers Who Bought All Products](./leetcode/medium/1045.%20Customers%20Who%20Bought%20All%20Products.sql)
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45 changes: 45 additions & 0 deletions leetcode/medium/585. Investments in 2016.sql
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/*
Question 585. Investments in 2016
Link: https://leetcode.com/problems/investments-in-2016/description/?envType=study-plan-v2&envId=top-sql-50

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.
*/

SELECT ROUND(SUM(i.tiv_2016)::numeric, 2) AS tiv_2016
FROM Insurance AS i
WHERE (i.lat, i.lon) IN (
SELECT
i1.lat,
i1.lon
FROM Insurance AS i1
GROUP BY i1.lat, i1.lon
HAVING COUNT(1) = 1
) AND i.tiv_2015 IN (
SELECT i2.tiv_2015
FROM Insurance AS i2
GROUP BY i2.tiv_2015
HAVING COUNT(1) > 1
)