diff --git a/LeetCode SQL 50 Solution/1321. Restaurant Growth.sql b/LeetCode SQL 50 Solution/1321. Restaurant Growth.sql
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+321. Restaurant Growth
+Solved
+Medium
+Topics
+Companies
+SQL Schema
+Pandas Schema
+Table: Customer
+
++---------------+---------+
+| Column Name   | Type    |
++---------------+---------+
+| customer_id   | int     |
+| name          | varchar |
+| visited_on    | date    |
+| amount        | int     |
++---------------+---------+
+In SQL,(customer_id, visited_on) is the primary key for this table.
+This table contains data about customer transactions in a restaurant.
+visited_on is the date on which the customer with ID (customer_id) has visited the restaurant.
+amount is the total paid by a customer.
+ 
+
+You are the restaurant owner and you want to analyze a possible expansion (there will be at least one customer every day).
+
+Compute the moving average of how much the customer paid in a seven days window (i.e., current day + 6 days before). average_amount should be rounded to two decimal places.
+
+Return the result table ordered by visited_on in ascending order.
+
+The result format is in the following example.
+
+ 
+
+Example 1:
+
+Input: 
+Customer table:
++-------------+--------------+--------------+-------------+
+| customer_id | name         | visited_on   | amount      |
++-------------+--------------+--------------+-------------+
+| 1           | Jhon         | 2019-01-01   | 100         |
+| 2           | Daniel       | 2019-01-02   | 110         |
+| 3           | Jade         | 2019-01-03   | 120         |
+| 4           | Khaled       | 2019-01-04   | 130         |
+| 5           | Winston      | 2019-01-05   | 110         | 
+| 6           | Elvis        | 2019-01-06   | 140         | 
+| 7           | Anna         | 2019-01-07   | 150         |
+| 8           | Maria        | 2019-01-08   | 80          |
+| 9           | Jaze         | 2019-01-09   | 110         | 
+| 1           | Jhon         | 2019-01-10   | 130         | 
+| 3           | Jade         | 2019-01-10   | 150         | 
++-------------+--------------+--------------+-------------+
+Output: 
++--------------+--------------+----------------+
+| visited_on   | amount       | average_amount |
++--------------+--------------+----------------+
+| 2019-01-07   | 860          | 122.86         |
+| 2019-01-08   | 840          | 120            |
+| 2019-01-09   | 840          | 120            |
+| 2019-01-10   | 1000         | 142.86         |
++--------------+--------------+----------------+
+Explanation: 
+1st moving average from 2019-01-01 to 2019-01-07 has an average_amount of (100 + 110 + 120 + 130 + 110 + 140 + 150)/7 = 122.86
+2nd moving average from 2019-01-02 to 2019-01-08 has an average_amount of (110 + 120 + 130 + 110 + 140 + 150 + 80)/7 = 120
+3rd moving average from 2019-01-03 to 2019-01-09 has an average_amount of (120 + 130 + 110 + 140 + 150 + 80 + 110)/7 = 120
+4th moving average from 2019-01-04 to 2019-01-10 has an average_amount of (130 + 110 + 140 + 150 + 80 + 110 + 130 + 150)/7 = 142.86
+
+
+Solution 1:
+
+# Write your MySQL query statement below
+WITH
+    t AS (
+        SELECT
+            visited_on,
+            SUM(amount) OVER (
+                ORDER BY visited_on
+                ROWS 6 PRECEDING
+            ) AS amount,
+            RANK() OVER (
+                ORDER BY visited_on
+                ROWS 6 PRECEDING
+            ) AS rk
+        FROM
+            (
+                SELECT visited_on, SUM(amount) AS amount
+                FROM Customer
+                GROUP BY visited_on
+            ) AS tt
+    )
+SELECT visited_on, amount, ROUND(amount / 7, 2) AS average_amount
+FROM t
+WHERE rk > 6;
+
+
+Solution 2:
+# Write your MySQL query statement below
+SELECT
+    a.visited_on,
+    SUM(b.amount) AS amount,
+    ROUND(SUM(b.amount) / 7, 2) AS average_amount
+FROM
+    (SELECT DISTINCT visited_on FROM customer) AS a
+    JOIN customer AS b ON DATEDIFF(a.visited_on, b.visited_on) BETWEEN 0 AND 6
+WHERE a.visited_on >= (SELECT MIN(visited_on) FROM customer) + 6
+GROUP BY 1
+ORDER BY 1;
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diff --git a/LeetCode SQL 50 Solution/1341. Movie Rating.sql b/LeetCode SQL 50 Solution/1341. Movie Rating.sql
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+1341. Movie Rating
+"""
+Table: Movies
+
++---------------+---------+
+| Column Name   | Type    |
++---------------+---------+
+| movie_id      | int     |
+| title         | varchar |
++---------------+---------+
+movie_id is the primary key (column with unique values) for this table.
+title is the name of the movie.
+ 
+
+Table: Users
+
++---------------+---------+
+| Column Name   | Type    |
++---------------+---------+
+| user_id       | int     |
+| name          | varchar |
++---------------+---------+
+user_id is the primary key (column with unique values) for this table.
+The column 'name' has unique values.
+Table: MovieRating
+
++---------------+---------+
+| Column Name   | Type    |
++---------------+---------+
+| movie_id      | int     |
+| user_id       | int     |
+| rating        | int     |
+| created_at    | date    |
++---------------+---------+
+(movie_id, user_id) is the primary key (column with unique values) for this table.
+This table contains the rating of a movie by a user in their review.
+created_at is the user's review date. 
+ 
+
+Write a solution to:
+
+Find the name of the user who has rated the greatest number of movies. In case of a tie, return the lexicographically smaller user name.
+Find the movie name with the highest average rating in February 2020. In case of a tie, return the lexicographically smaller movie name.
+The result format is in the following example.
+
+ 
+
+Example 1:
+
+Input: 
+Movies table:
++-------------+--------------+
+| movie_id    |  title       |
++-------------+--------------+
+| 1           | Avengers     |
+| 2           | Frozen 2     |
+| 3           | Joker        |
++-------------+--------------+
+Users table:
++-------------+--------------+
+| user_id     |  name        |
++-------------+--------------+
+| 1           | Daniel       |
+| 2           | Monica       |
+| 3           | Maria        |
+| 4           | James        |
++-------------+--------------+
+MovieRating table:
++-------------+--------------+--------------+-------------+
+| movie_id    | user_id      | rating       | created_at  |
++-------------+--------------+--------------+-------------+
+| 1           | 1            | 3            | 2020-01-12  |
+| 1           | 2            | 4            | 2020-02-11  |
+| 1           | 3            | 2            | 2020-02-12  |
+| 1           | 4            | 1            | 2020-01-01  |
+| 2           | 1            | 5            | 2020-02-17  | 
+| 2           | 2            | 2            | 2020-02-01  | 
+| 2           | 3            | 2            | 2020-03-01  |
+| 3           | 1            | 3            | 2020-02-22  | 
+| 3           | 2            | 4            | 2020-02-25  | 
++-------------+--------------+--------------+-------------+
+Output: 
++--------------+
+| results      |
++--------------+
+| Daniel       |
+| Frozen 2     |
++--------------+
+Explanation: 
+Daniel and Monica have rated 3 movies ("Avengers", "Frozen 2" and "Joker") but Daniel is smaller lexicographically.
+Frozen 2 and Joker have a rating average of 3.5 in February but Frozen 2 is smaller lexicographically.
+
+"""
+
+# Write your MySQL query statement below
+(
+    SELECT name AS results
+    FROM
+        Users
+        JOIN MovieRating USING (user_id)
+    GROUP BY user_id
+    ORDER BY COUNT(1) DESC, name
+    LIMIT 1
+)
+UNION ALL
+(
+    SELECT title
+    FROM
+        MovieRating
+        JOIN Movies USING (movie_id)
+    WHERE DATE_FORMAT(created_at, '%Y-%m') = '2020-02'
+    GROUP BY movie_id
+    ORDER BY AVG(rating) DESC, title
+    LIMIT 1
+);
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