task: #1321

Ivanbyone · Ivanbyone · commit f2196e651f0d · 2025-07-09T12:41:54.000+03:00
diff --git a/README.md b/README.md
@@ -92,6 +92,7 @@ Have a good contributing!
     - [1174. Immediate Food Delivery II](./leetcode/medium/1174.%20Immediate%20Food%20Delivery%20II.sql)
     - [1193. Monthly Transactions I](./leetcode/medium/1193.%20Monthly%20Transactions%20I.sql)
     - [1204. Last Person to Fit in the Bus](./leetcode/medium/1204.%20Last%20Person%20to%20Fit%20in%20the%20Bus.sql)
+    - [1321. Restaurant Growth](./leetcode/medium/1321.%20Restaurant%20Growth.sql)
     - [1341. Movie Rating](./leetcode/medium/1341.%20Movie%20Rating.sql)
     - [1907. Count Salary Categories](./leetcode/medium/1907.%20Count%20Salary%20Categories.sql)
     - [1934. Confirmation Rate](./leetcode/medium/1934.%20Confirmation%20Rate.sql)
diff --git a/leetcode/medium/1321. Restaurant Growth.sql b/leetcode/medium/1321. Restaurant Growth.sql
@@ -0,0 +1,50 @@
+/*
+Question 1321. Restaurant Growth
+Link: https://leetcode.com/problems/restaurant-growth/description/?envType=study-plan-v2&envId=top-sql-50
+
+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.
+*/
+
+WITH visited_dates AS (
+    SELECT visited_on + INTERVAL '6 days' AS dates
+    FROM Customer
+    GROUP BY visited_on
+)
+
+SELECT DISTINCT
+    c.visited_on,
+    (
+        SELECT SUM(c1.amount)
+        FROM Customer AS c1
+        WHERE c1.visited_on BETWEEN c.visited_on - INTERVAL '6 days' AND c.visited_on
+    ) AS amount,
+    (
+        SELECT ROUND(SUM(c2.amount)::NUMERIC / 7, 2)
+        FROM Customer AS c2
+        WHERE c2.visited_on BETWEEN c.visited_on - INTERVAL '6 days' AND c.visited_on
+    ) AS average_amount
+FROM Customer AS c
+LEFT JOIN
+    visited_dates AS v
+    ON c.visited_on = v.dates
+WHERE v.dates IS NOT NULL
