LeetCode #1321 — MEDIUM

Restaurant Growth

Move from brute-force thinking to an efficient approach using core interview patterns strategy.

The Problem

Problem Statement

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: General problem-solving

Example 1

{"headers":{"Customer":["customer_id","name","visited_on","amount"]},"rows":{"Customer":[[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]]}}

Step 02

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1321: Restaurant Growth
// Auto-generated Java example from rust.
class Solution {
    public void exampleSolution() {
    }
}
// Reference (rust):
// // Accepted solution for LeetCode #1321: Restaurant Growth
// pub fn sql_example() -> &'static str {
//     r#"
// -- Accepted solution for LeetCode #1321: Restaurant Growth
// # 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;
// "#
// }

// Accepted solution for LeetCode #1321: Restaurant Growth
// Auto-generated Go example from rust.
func exampleSolution() {
}
// Reference (rust):
// // Accepted solution for LeetCode #1321: Restaurant Growth
// pub fn sql_example() -> &'static str {
//     r#"
// -- Accepted solution for LeetCode #1321: Restaurant Growth
// # 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;
// "#
// }

# Accepted solution for LeetCode #1321: Restaurant Growth
# Auto-generated Python example from rust.
def example_solution() -> None:
    return
# Reference (rust):
# // Accepted solution for LeetCode #1321: Restaurant Growth
# pub fn sql_example() -> &'static str {
#     r#"
# -- Accepted solution for LeetCode #1321: Restaurant Growth
# # 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;
# "#
# }

// Accepted solution for LeetCode #1321: Restaurant Growth
pub fn sql_example() -> &'static str {
    r#"
-- Accepted solution for LeetCode #1321: Restaurant Growth
# 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;
"#
}

// Accepted solution for LeetCode #1321: Restaurant Growth
// Auto-generated TypeScript example from rust.
function exampleSolution(): void {
}
// Reference (rust):
// // Accepted solution for LeetCode #1321: Restaurant Growth
// pub fn sql_example() -> &'static str {
//     r#"
// -- Accepted solution for LeetCode #1321: Restaurant Growth
// # 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;
// "#
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.