task: #1890

Ivanbyone · Ivanbyone · commit 7516dee7203e · 2025-07-10T11:07:35.000+03:00
diff --git a/README.md b/README.md
@@ -145,6 +145,7 @@ Useful for preparing for technical interviews and improving your SQL skills.
     - [1741. Find Total Time Spent by Each Employee](./leetcode/easy/1741.%20Find%20Total%20Time%20Spent%20by%20Each%20Employee.sql)
     - [1789. Primary Department for Each Employee](./leetcode/easy/1789.%20Primary%20Department%20for%20Each%20Employee.sql)
     - [1795. Rearrange Products Table](./leetcode/easy/1795.%20Rearrange%20Products%20Table.sql)
+    - [1890. The Latest Login in 2020](./leetcode/easy/1890.%20The%20Latest%20Login%20in%202020.sql)
     - [1965. Employees With Missing Information](./leetcode/easy/1965.%20Employees%20With%20Missing%20Information.sql)
     - [1978. Employees Whose Manager Left the Company](./leetcode/easy/1978.%20Employees%20Whose%20Manager%20Left%20the%20Company.sql)
     - [2356. Number of Unique Subjects Taught by Each Teacher](./leetcode/easy/2356.%20Number%20of%20Unique%20Subjects%20Taught%20by%20Each%20Teacher.sql)
diff --git a/leetcode/easy/1890. The Latest Login in 2020.sql b/leetcode/easy/1890. The Latest Login in 2020.sql
@@ -0,0 +1,27 @@
+/*
+Question 1890. The Latest Login in 2020
+Link: https://leetcode.com/problems/the-latest-login-in-2020/description/?envType=problem-list-v2&envId=database
+
+Table: Logins
+
++----------------+----------+
+| Column Name    | Type     |
++----------------+----------+
+| user_id        | int      |
+| time_stamp     | datetime |
++----------------+----------+
+(user_id, time_stamp) is the primary key (combination of columns with unique values) for this table.
+Each row contains information about the login time for the user with ID user_id.
+ 
+
+Write a solution to report the latest login for all users in the year 2020. Do not include the users who did not login in 2020.
+
+Return the result table in any order.
+*/
+
+SELECT
+    user_id,
+    MAX(time_stamp) AS last_stamp
+FROM Logins
+WHERE time_stamp::varchar LIKE '2020%'
+GROUP BY user_id
