diff --git a/LeetCode SQL 50 Solution/585. Investments in 2016.sql b/LeetCode SQL 50 Solution/585. Investments in 2016.sql
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+585. Investments in 2016
+"""
+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.
+
+The result format is in the following example.
+
+ 
+
+Example 1:
+
+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.
+
+"""
+
+WITH
+  InsuranceWithCounts AS (
+    SELECT
+      tiv_2016,
+      COUNT(*) OVER(PARTITION by tiv_2015) AS tiv_2015_count,
+      COUNT(*) OVER(PARTITION by lat, lon) AS city_count
+    FROM Insurance
+  )
+SELECT ROUND(SUM(tiv_2016), 2) AS tiv_2016
+FROM InsuranceWithCounts
+WHERE tiv_2015_count > 1
+  AND city_count = 1;
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