diff --git a/LeetCode SQL 50 Solution/602. Friend Requests II.sql b/LeetCode SQL 50 Solution/602. Friend Requests II.sql
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+602. Friend Requests II: Who Has the Most Friends
+Solved
+Medium
+Topics
+Companies
+Hint
+SQL Schema
+Pandas Schema
+Table: RequestAccepted
+
++----------------+---------+
+| Column Name    | Type    |
++----------------+---------+
+| requester_id   | int     |
+| accepter_id    | int     |
+| accept_date    | date    |
++----------------+---------+
+(requester_id, accepter_id) is the primary key (combination of columns with unique values) for this table.
+This table contains the ID of the user who sent the request, the ID of the user who received the request, and the date when the request was accepted.
+ 
+
+Write a solution to find the people who have the most friends and the most friends number.
+
+The test cases are generated so that only one person has the most friends.
+
+The result format is in the following example.
+
+ 
+
+Example 1:
+
+Input: 
+RequestAccepted table:
++--------------+-------------+-------------+
+| requester_id | accepter_id | accept_date |
++--------------+-------------+-------------+
+| 1            | 2           | 2016/06/03  |
+| 1            | 3           | 2016/06/08  |
+| 2            | 3           | 2016/06/08  |
+| 3            | 4           | 2016/06/09  |
++--------------+-------------+-------------+
+Output: 
++----+-----+
+| id | num |
++----+-----+
+| 3  | 3   |
++----+-----+
+Explanation: 
+The person with id 3 is a friend of people 1, 2, and 4, so he has three friends in total, which is the most number than any others.
+ 
+
+Follow up: In the real world, multiple people could have the same most number of friends. Could you find all these people in this case?
+
+
+
+# Write your MySQL query statement below
+WITH
+    T AS (
+        SELECT requester_id, accepter_id FROM RequestAccepted
+        UNION ALL
+        SELECT accepter_id, requester_id FROM RequestAccepted
+    )
+SELECT requester_id AS id, COUNT(1) AS num
+FROM T
+GROUP BY 1
+ORDER BY 2 DESC
+LIMIT 1;
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diff --git a/LeetCode SQL 50 Solution/626. Exchange Seats.sql b/LeetCode SQL 50 Solution/626. Exchange Seats.sql
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+++ b/LeetCode SQL 50 Solution/626. Exchange Seats.sql	
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+626. Exchange Seats
+Solved
+Medium
+Topics
+Companies
+SQL Schema
+Pandas Schema
+Table: Seat
+
++-------------+---------+
+| Column Name | Type    |
++-------------+---------+
+| id          | int     |
+| student     | varchar |
++-------------+---------+
+id is the primary key (unique value) column for this table.
+Each row of this table indicates the name and the ID of a student.
+The ID sequence always starts from 1 and increments continuously.
+ 
+
+Write a solution to swap the seat id of every two consecutive students. If the number of students is odd, the id of the last student is not swapped.
+
+Return the result table ordered by id in ascending order.
+
+The result format is in the following example.
+
+ 
+
+Example 1:
+
+Input: 
+Seat table:
++----+---------+
+| id | student |
++----+---------+
+| 1  | Abbot   |
+| 2  | Doris   |
+| 3  | Emerson |
+| 4  | Green   |
+| 5  | Jeames  |
++----+---------+
+Output: 
++----+---------+
+| id | student |
++----+---------+
+| 1  | Doris   |
+| 2  | Abbot   |
+| 3  | Green   |
+| 4  | Emerson |
+| 5  | Jeames  |
++----+---------+
+Explanation: 
+Note that if the number of students is odd, there is no need to change the last one's seat.
+'''
+
+# MYSQL Query Accepted
+
+SELECT ( CASE
+            WHEN id%2 != 0 AND id != counts THEN id+1
+            WHEN id%2 != 0 AND id = counts THEN id
+            ELSE id-1
+        END) AS id, student
+FROM seat, (select count(*) as counts from seat) 
+AS seat_counts
+ORDER BY id ASC
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