Solved -> 1321. Restaurant Growth

Author: NandiSoham

06_Subqueries/1321. Restaurant Growth.sql

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+/* 
+
+Problem Link -> https://leetcode.com/problems/restaurant-growth/description/?envType=study-plan-v2&envId=top-sql-50
+
+------------------------------------------------------------- QUESTION -----------------------------------------------------------
+
+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 -----------------------------------------------------------
+
+SELECT
+    visited_on,
+    (
+        SELECT SUM(amount)
+        FROM Customer
+        WHERE visited_on BETWEEN DATE_SUB(c.visited_on, INTERVAL 6 DAY) AND c.visited_on
+    ) AS amount,
+    ROUND((
+        SELECT SUM(amount)/7
+        FROM Customer
+        WHERE visited_on BETWEEN DATE_SUB(c.visited_on, INTERVAL 6 DAY) AND c.visited_on
+    ),2) AS average_amount
+FROM Customer c
+WHERE visited_on >= (
+    SELECT DATE_ADD(MIN(visited_on), INTERVAL 6 DAY)
+    FROM Customer
+)
+GROUP BY visited_on
+ORDER BY visited_on;
+