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#1321
Medium Database

Restaurant growth

Database
58.2% acceptance
Feb 27, 2026
1149
380
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.

Solution

SQL
LeetCode
solution.sql
#
# 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
#

# Write your MySQL query statement below

SELECT visited_on,
  SUM(amount) OVER (ORDER BY visited_on ROWS BETWEEN 6 PRECEDING AND CURRENT ROW) AS amount,
  ROUND(AVG(amount) OVER (ORDER BY visited_on ROWS BETWEEN 6 PRECEDING AND CURRENT ROW), 2) AS average_amount
FROM (
  SELECT visited_on, SUM(amount) AS amount
  FROM Customer
  GROUP BY visited_on
) daily
ORDER BY visited_on
LIMIT 18446744073709551615 OFFSET 6;