LeetCode #3705 — MEDIUM

Find Golden Hour Customers

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

The Problem

Problem Statement

Table: restaurant_orders

+------------------+----------+
| Column Name      | Type     | 
+------------------+----------+
| order_id         | int      |
| customer_id      | int      |
| order_timestamp  | datetime |
| order_amount     | decimal  |
| payment_method   | varchar  |
| order_rating     | int      |
+------------------+----------+
order_id is the unique identifier for this table.
payment_method can be cash, card, or app.
order_rating is between 1 and 5, where 5 is the best (NULL if not rated).
order_timestamp contains both date and time information.

Write a solution to find golden hour customers - customers who consistently order during peak hours and provide high satisfaction. A customer is a golden hour customer if they meet ALL the following criteria:

Made at least 3 orders.
At least 60% of their orders are during peak hours (11:00-14:00 or 18:00-21:00).
Their average rating for rated orders is at least 4.0, round it to 2 decimal places.
Have rated at least 50% of their orders.

Return the result table ordered by average_rating in descending order, then by customer_id in descending order.

The result format is in the following example.

Example:

Input:

restaurant_orders table:

+----------+-------------+---------------------+--------------+----------------+--------------+
| order_id | customer_id | order_timestamp     | order_amount | payment_method | order_rating |
+----------+-------------+---------------------+--------------+----------------+--------------+
| 1        | 101         | 2024-03-01 12:30:00 | 25.50        | card           | 5            |
| 2        | 101         | 2024-03-02 19:15:00 | 32.00        | app            | 4            |
| 3        | 101         | 2024-03-03 13:45:00 | 28.75        | card           | 5            |
| 4        | 101         | 2024-03-04 20:30:00 | 41.00        | app            | NULL         |
| 5        | 102         | 2024-03-01 11:30:00 | 18.50        | cash           | 4            |
| 6        | 102         | 2024-03-02 12:00:00 | 22.00        | card           | 3            |
| 7        | 102         | 2024-03-03 15:30:00 | 19.75        | cash           | NULL         |
| 8        | 103         | 2024-03-01 19:00:00 | 55.00        | app            | 5            |
| 9        | 103         | 2024-03-02 20:45:00 | 48.50        | app            | 4            |
| 10       | 103         | 2024-03-03 18:30:00 | 62.00        | card           | 5            |
| 11       | 104         | 2024-03-01 10:00:00 | 15.00        | cash           | 3            |
| 12       | 104         | 2024-03-02 09:30:00 | 18.00        | cash           | 2            |
| 13       | 104         | 2024-03-03 16:00:00 | 20.00        | card           | 3            |
| 14       | 105         | 2024-03-01 12:15:00 | 30.00        | app            | 4            |
| 15       | 105         | 2024-03-02 13:00:00 | 35.50        | app            | 5            |
| 16       | 105         | 2024-03-03 11:45:00 | 28.00        | card           | 4            |
+----------+-------------+---------------------+--------------+----------------+--------------+

Output:

+-------------+--------------+----------------------+----------------+
| customer_id | total_orders | peak_hour_percentage | average_rating |
+-------------+--------------+----------------------+----------------+
| 103         | 3            | 100                  | 4.67           |
| 101         | 4            | 100                  | 4.67           |
| 105         | 3            | 100                  | 4.33           |
+-------------+--------------+----------------------+----------------+

Explanation:

Customer 101:
- Total orders: 4 (at least 3)
- Peak hour orders: 4 out of 4 (12:30, 19:15, 13:45, and 20:30 are in peak hours)
- Peak hour percentage: 100% (at least 60%)
- Rated orders: 3 out of 4 (75% rating completion)
- Average rating: (5+4+5)/3 = 4.67 (at least 4.0)
- Result: Golden hour customer
Customer 102:
- Total orders: 3 (at least 3)
- Peak hour orders: 2 out of 3 (11:30, 12:00 are in peak hours; 15:30 is not)
- Peak hour percentage: 2/3 = 66.67% (at least 60%)
- Rated orders: 2 out of 3 (66.67% rating completion)
- Average rating: (4+3)/2 = 3.5 (less than 4.0)
- Result: Not a golden hour customer (average rating too low)
Customer 103:
- Total orders: 3 (at least 3)
- Peak hour orders: 3 out of 3 (19:00, 20:45, 18:30 all in evening peak)
- Peak hour percentage: 3/3 = 100% (at least 60%)
- Rated orders: 3 out of 3 (100% rating completion)
- Average rating: (5+4+5)/3 = 4.67 (at least 4.0)
- Result: Golden hour customer
Customer 104:
- Total orders: 3 (at least 3)
- Peak hour orders: 0 out of 3 (10:00, 09:30, 16:00 all outside peak hours)
- Peak hour percentage: 0/3 = 0% (less than 60%)
- Result: Not a golden hour customer (insufficient peak hour orders)
Customer 105:
- Total orders: 3 (at least 3)
- Peak hour orders: 3 out of 3 (12:15, 13:00, 11:45 all in lunch peak)
- Peak hour percentage: 3/3 = 100% (at least 60%)
- Rated orders: 3 out of 3 (100% rating completion)
- Average rating: (4+5+4)/3 = 4.33 (at least 4.0)
- Result: Golden hour customer

The results table is ordered by average_rating DESC, then customer_id DESC.

Step 01

Brute Force Baseline

Problem summary: Table: restaurant_orders +------------------+----------+ | Column Name | Type | +------------------+----------+ | order_id | int | | customer_id | int | | order_timestamp | datetime | | order_amount | decimal | | payment_method | varchar | | order_rating | int | +------------------+----------+ order_id is the unique identifier for this table. payment_method can be cash, card, or app. order_rating is between 1 and 5, where 5 is the best (NULL if not rated). order_timestamp contains both date and time information. Write a solution to find golden hour customers - customers who consistently order during peak hours and provide high satisfaction. A customer is a golden hour customer if they meet ALL the following criteria: Made at least 3 orders. At least 60% of their orders are during peak hours (11:00-14:00 or 18:00-21:00). Their average rating for rated orders is at least 4.0, round it to

Baseline thinking

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

Pattern signal: General problem-solving

Example 1

{"headers":{"restaurant_orders":["order_id","customer_id","order_timestamp","order_amount","payment_method","order_rating"]},"rows":{"restaurant_orders":[[1,101,"2024-03-01 12:30:00",25.50,"card",5],[2,101,"2024-03-02 19:15:00",32.00,"app",4],[3,101,"2024-03-03 13:45:00",28.75,"card",5],[4,101,"2024-03-04 20:30:00",41.00,"app",null],[5,102,"2024-03-01 11:30:00",18.50,"cash",4],[6,102,"2024-03-02 12:00:00",22.00,"card",3],[7,102,"2024-03-03 15:30:00",19.75,"cash",null],[8,103,"2024-03-01 19:00:00",55.00,"app",5],[9,103,"2024-03-02 20:45:00",48.50,"app",4],[10,103,"2024-03-03 18:30:00",62.00,"card",5],[11,104,"2024-03-01 10:00:00",15.00,"cash",3],[12,104,"2024-03-02 09:30:00",18.00,"cash",2],[13,104,"2024-03-03 16:00:00",20.00,"card",3],[14,105,"2024-03-01 12:15:00",30.00,"app",4],[15,105,"2024-03-02 13:00:00",35.50,"app",5],[16,105,"2024-03-03 11:45:00",28.00,"card",4]]}}

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 #3705: Find Golden Hour Customers
// Auto-generated Java example from py.
class Solution {
    public void exampleSolution() {
    }
}
// Reference (py):
// # Accepted solution for LeetCode #3705: Find Golden Hour Customers
// import pandas as pd
// import numpy as np
// 
// 
// def find_golden_hour_customers(restaurant_orders: pd.DataFrame) -> pd.DataFrame:
//     df = restaurant_orders.copy()
//     df["order_timestamp"] = pd.to_datetime(df["order_timestamp"])
//     df["is_peak_hour"] = df["order_timestamp"].dt.time.between(
//         pd.to_datetime("11:00:00").time(), pd.to_datetime("14:00:00").time()
//     ) | df["order_timestamp"].dt.time.between(
//         pd.to_datetime("18:00:00").time(), pd.to_datetime("21:00:00").time()
//     )
//     grouped = (
//         df.groupby("customer_id")
//         .agg(
//             total_orders=("order_timestamp", "count"),
//             peak_hour_count=("is_peak_hour", "sum"),
//             average_rating=("order_rating", lambda x: x.dropna().mean()),
//             non_null_rating_count=("order_rating", lambda x: x.notna().sum()),
//         )
//         .reset_index()
//     )
//     grouped["average_rating"] = grouped["average_rating"].round(2)
//     grouped["peak_hour_percentage"] = (
//         grouped["peak_hour_count"] / grouped["total_orders"] * 100
//     ).round()
//     filtered = grouped[
//         (grouped["total_orders"] >= 3)
//         & (grouped["peak_hour_percentage"] >= 60)
//         & (grouped["average_rating"] >= 4.0)
//         & (grouped["non_null_rating_count"] / grouped["total_orders"] >= 0.5)
//     ]
//     filtered = filtered.sort_values(
//         by=["average_rating", "customer_id"], ascending=[False, False]
//     )
//     return filtered[
//         ["customer_id", "total_orders", "peak_hour_percentage", "average_rating"]
//     ]

// Accepted solution for LeetCode #3705: Find Golden Hour Customers
// Auto-generated Go example from py.
func exampleSolution() {
}
// Reference (py):
// # Accepted solution for LeetCode #3705: Find Golden Hour Customers
// import pandas as pd
// import numpy as np
// 
// 
// def find_golden_hour_customers(restaurant_orders: pd.DataFrame) -> pd.DataFrame:
//     df = restaurant_orders.copy()
//     df["order_timestamp"] = pd.to_datetime(df["order_timestamp"])
//     df["is_peak_hour"] = df["order_timestamp"].dt.time.between(
//         pd.to_datetime("11:00:00").time(), pd.to_datetime("14:00:00").time()
//     ) | df["order_timestamp"].dt.time.between(
//         pd.to_datetime("18:00:00").time(), pd.to_datetime("21:00:00").time()
//     )
//     grouped = (
//         df.groupby("customer_id")
//         .agg(
//             total_orders=("order_timestamp", "count"),
//             peak_hour_count=("is_peak_hour", "sum"),
//             average_rating=("order_rating", lambda x: x.dropna().mean()),
//             non_null_rating_count=("order_rating", lambda x: x.notna().sum()),
//         )
//         .reset_index()
//     )
//     grouped["average_rating"] = grouped["average_rating"].round(2)
//     grouped["peak_hour_percentage"] = (
//         grouped["peak_hour_count"] / grouped["total_orders"] * 100
//     ).round()
//     filtered = grouped[
//         (grouped["total_orders"] >= 3)
//         & (grouped["peak_hour_percentage"] >= 60)
//         & (grouped["average_rating"] >= 4.0)
//         & (grouped["non_null_rating_count"] / grouped["total_orders"] >= 0.5)
//     ]
//     filtered = filtered.sort_values(
//         by=["average_rating", "customer_id"], ascending=[False, False]
//     )
//     return filtered[
//         ["customer_id", "total_orders", "peak_hour_percentage", "average_rating"]
//     ]

# Accepted solution for LeetCode #3705: Find Golden Hour Customers
import pandas as pd
import numpy as np


def find_golden_hour_customers(restaurant_orders: pd.DataFrame) -> pd.DataFrame:
    df = restaurant_orders.copy()
    df["order_timestamp"] = pd.to_datetime(df["order_timestamp"])
    df["is_peak_hour"] = df["order_timestamp"].dt.time.between(
        pd.to_datetime("11:00:00").time(), pd.to_datetime("14:00:00").time()
    ) | df["order_timestamp"].dt.time.between(
        pd.to_datetime("18:00:00").time(), pd.to_datetime("21:00:00").time()
    )
    grouped = (
        df.groupby("customer_id")
        .agg(
            total_orders=("order_timestamp", "count"),
            peak_hour_count=("is_peak_hour", "sum"),
            average_rating=("order_rating", lambda x: x.dropna().mean()),
            non_null_rating_count=("order_rating", lambda x: x.notna().sum()),
        )
        .reset_index()
    )
    grouped["average_rating"] = grouped["average_rating"].round(2)
    grouped["peak_hour_percentage"] = (
        grouped["peak_hour_count"] / grouped["total_orders"] * 100
    ).round()
    filtered = grouped[
        (grouped["total_orders"] >= 3)
        & (grouped["peak_hour_percentage"] >= 60)
        & (grouped["average_rating"] >= 4.0)
        & (grouped["non_null_rating_count"] / grouped["total_orders"] >= 0.5)
    ]
    filtered = filtered.sort_values(
        by=["average_rating", "customer_id"], ascending=[False, False]
    )
    return filtered[
        ["customer_id", "total_orders", "peak_hour_percentage", "average_rating"]
    ]

// Accepted solution for LeetCode #3705: Find Golden Hour Customers
// Rust example auto-generated from py reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (py):
// # Accepted solution for LeetCode #3705: Find Golden Hour Customers
// import pandas as pd
// import numpy as np
// 
// 
// def find_golden_hour_customers(restaurant_orders: pd.DataFrame) -> pd.DataFrame:
//     df = restaurant_orders.copy()
//     df["order_timestamp"] = pd.to_datetime(df["order_timestamp"])
//     df["is_peak_hour"] = df["order_timestamp"].dt.time.between(
//         pd.to_datetime("11:00:00").time(), pd.to_datetime("14:00:00").time()
//     ) | df["order_timestamp"].dt.time.between(
//         pd.to_datetime("18:00:00").time(), pd.to_datetime("21:00:00").time()
//     )
//     grouped = (
//         df.groupby("customer_id")
//         .agg(
//             total_orders=("order_timestamp", "count"),
//             peak_hour_count=("is_peak_hour", "sum"),
//             average_rating=("order_rating", lambda x: x.dropna().mean()),
//             non_null_rating_count=("order_rating", lambda x: x.notna().sum()),
//         )
//         .reset_index()
//     )
//     grouped["average_rating"] = grouped["average_rating"].round(2)
//     grouped["peak_hour_percentage"] = (
//         grouped["peak_hour_count"] / grouped["total_orders"] * 100
//     ).round()
//     filtered = grouped[
//         (grouped["total_orders"] >= 3)
//         & (grouped["peak_hour_percentage"] >= 60)
//         & (grouped["average_rating"] >= 4.0)
//         & (grouped["non_null_rating_count"] / grouped["total_orders"] >= 0.5)
//     ]
//     filtered = filtered.sort_values(
//         by=["average_rating", "customer_id"], ascending=[False, False]
//     )
//     return filtered[
//         ["customer_id", "total_orders", "peak_hour_percentage", "average_rating"]
//     ]

// Accepted solution for LeetCode #3705: Find Golden Hour Customers
// Auto-generated TypeScript example from py.
function exampleSolution(): void {
}
// Reference (py):
// # Accepted solution for LeetCode #3705: Find Golden Hour Customers
// import pandas as pd
// import numpy as np
// 
// 
// def find_golden_hour_customers(restaurant_orders: pd.DataFrame) -> pd.DataFrame:
//     df = restaurant_orders.copy()
//     df["order_timestamp"] = pd.to_datetime(df["order_timestamp"])
//     df["is_peak_hour"] = df["order_timestamp"].dt.time.between(
//         pd.to_datetime("11:00:00").time(), pd.to_datetime("14:00:00").time()
//     ) | df["order_timestamp"].dt.time.between(
//         pd.to_datetime("18:00:00").time(), pd.to_datetime("21:00:00").time()
//     )
//     grouped = (
//         df.groupby("customer_id")
//         .agg(
//             total_orders=("order_timestamp", "count"),
//             peak_hour_count=("is_peak_hour", "sum"),
//             average_rating=("order_rating", lambda x: x.dropna().mean()),
//             non_null_rating_count=("order_rating", lambda x: x.notna().sum()),
//         )
//         .reset_index()
//     )
//     grouped["average_rating"] = grouped["average_rating"].round(2)
//     grouped["peak_hour_percentage"] = (
//         grouped["peak_hour_count"] / grouped["total_orders"] * 100
//     ).round()
//     filtered = grouped[
//         (grouped["total_orders"] >= 3)
//         & (grouped["peak_hour_percentage"] >= 60)
//         & (grouped["average_rating"] >= 4.0)
//         & (grouped["non_null_rating_count"] / grouped["total_orders"] >= 0.5)
//     ]
//     filtered = filtered.sort_values(
//         by=["average_rating", "customer_id"], ascending=[False, False]
//     )
//     return filtered[
//         ["customer_id", "total_orders", "peak_hour_percentage", "average_rating"]
//     ]

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.