Hopper Company Queries II - Practice Coding Problems

Hopper Company Queries II - Problem

Database Hard

Hopper Company Queries II is a challenging SQL problem that involves calculating the percentage of active drivers for each month in 2020.

You're given three tables:

Drivers: Contains driver information and their join dates
Rides: Contains all ride requests (both accepted and rejected)
AcceptedRides: Contains details about rides that were actually completed

Your task is to calculate the working percentage for each month, which is defined as:

(Number of drivers who accepted at least one ride during the month) / (Total number of available drivers during the month) × 100

A driver is considered available in a month if they had already joined the company by that month. A driver is considered working if they accepted at least one ride during that specific month.

Key Requirements:

Report results for all 12 months of 2020
Round the working percentage to 2 decimal places
If no drivers are available in a month, return 0.00%
Order results by month (1=January, 2=February, etc.)

Input & Output

example_1.sql — Basic Example

$ Input: Drivers: [(1,'2019-01-01'), (2,'2019-03-01'), (3,'2020-01-01')] Rides: [(1,1,'2020-01-15'), (2,2,'2020-02-14'), (3,3,'2020-03-15')] AcceptedRides: [(1,1,20,30), (3,3,15,25)]

› Output: [(1, 33.33), (2, 0.00), (3, 50.00), (4, 0.00), ..., (12, 0.00)]

💡 Note: January: 3 drivers available (all joined by Jan 31), 1 working (driver 1) → 1/3 = 33.33%. February: 3 available, 0 working → 0%. March: 3 available, 1 working (driver 3) → 33.33%

example_2.sql — No Available Drivers

$ Input: Drivers: [(1,'2020-02-01'), (2,'2020-03-01')] Rides: [(1,1,'2020-01-15')] AcceptedRides: [(1,1,10,20)]

› Output: [(1, 0.00), (2, 0.00), (3, 0.00), ..., (12, 0.00)]

💡 Note: January: 0 drivers available (none joined yet), so working_percentage = 0.00% by definition. February onwards: drivers become available but may not be working

example_3.sql — All Drivers Working

$ Input: Drivers: [(1,'2019-12-01'), (2,'2019-12-01')] Rides: [(1,1,'2020-06-15'), (2,2,'2020-06-16')] AcceptedRides: [(1,1,25,35), (2,2,30,40)]

› Output: [(1, 0.00), (2, 0.00), ..., (6, 100.00), (7, 0.00), ..., (12, 0.00)]

💡 Note: June: 2 drivers available (both joined before 2020), 2 working (both accepted rides) → 2/2 = 100.00%. All other months have 0 working drivers

Visualization

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Understanding the Visualization

Driver Timeline

Track when each driver joins the company (availability accumulates)

Monthly Activity

For each month, count drivers who accepted at least one ride

Calculate Ratios

Working drivers ÷ Available drivers × 100 for each month

Key Takeaway

🎯 Key Insight: Driver availability is cumulative (once joined, always available), while working status resets each month and requires actual ride activity.

Time & Space Complexity

Time Complexity

⏱️

O(n log n)

Single pass through data with efficient joins and window functions

⚡ Linearithmic

Space Complexity

O(n)

Space for intermediate CTE results and final output

⚡ Linearithmic Space

Constraints

1 ≤ drivers.length ≤ 10⁴
1 ≤ rides.length ≤ 10⁴
1 ≤ acceptedRides.length ≤ rides.length
All dates are valid and in proper format
Each driver_id is unique in Drivers table
Each ride_id is unique in Rides and AcceptedRides tables
AcceptedRides.ride_id always exists in Rides table

Asked in

U Uber 85 L Lyft 60 D DoorDash 45 a Amazon 30

The optimal solution uses Common Table Expressions (CTEs) with window functions to efficiently calculate driver availability and activity in a single query. The key insight is to generate all 12 months, calculate cumulative available drivers using date comparisons, and join with monthly working driver statistics to compute the final percentage.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force (Multiple Subqueries)	O(12 × n)	O(n)	Calculate available and working drivers separately for each month using multiple subqueries
CTE with Window Functions (Optimal)	O(n log n)	O(n)	Use CTEs and window functions to calculate running totals efficiently in a single pass

Brute Force (Multiple Subqueries) — Algorithm Steps

Create a query for each month (1-12) with separate subqueries
For each month, count available drivers (joined before or during that month)
For each month, count working drivers (accepted rides during that month)
Calculate percentage and round to 2 decimal places
Union all monthly results and order by month

Visualization

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Step-by-Step Walkthrough

January Calculation

Count available drivers (joined by Jan 31) and working drivers (active in Jan)

February Calculation

Repeat same process for February with separate queries

Continue for All Months

Repeat process 12 times, once for each month

Code -

solution.c — C

// C implementation with SQL string building
#include <stdio.h>
#include <string.h>

char* buildBruteForceQuery() {
    static char query[4096];
    char* monthEnds[] = {
        "2020-01-31", "2020-02-29", "2020-03-31", "2020-04-30",
        "2020-05-31", "2020-06-30", "2020-07-31", "2020-08-31",
        "2020-09-30", "2020-10-31", "2020-11-30", "2020-12-31"
    };
    
    strcpy(query, "");
    for (int month = 1; month <= 12; month++) {
        char monthQuery[512];
        if (month > 1) strcat(query, " UNION ALL ");
        sprintf(monthQuery, 
            "SELECT %d as month, "
            "ROUND(CASE WHEN available = 0 THEN 0.00 "
            "ELSE working * 100.0 / available END, 2) as working_percentage "
            "FROM (SELECT "
            "(SELECT COUNT(DISTINCT driver_id) FROM Drivers WHERE join_date <= '%s') as available, "
            "(SELECT COUNT(DISTINCT ar.driver_id) FROM AcceptedRides ar JOIN Rides r ON ar.ride_id = r.ride_id "
            "WHERE YEAR(r.requested_at) = 2020 AND MONTH(r.requested_at) = %d) as working) temp",
            month, monthEnds[month-1], month);
        strcat(query, monthQuery);
    }
    strcat(query, " ORDER BY month");
    return query;
}

Time & Space Complexity

Time Complexity

⏱️

O(12 × n)

12 months × n records, with multiple table scans per month

✓ Linear Growth

Space Complexity

O(n)

Space for intermediate results and final output

⚡ Linearithmic Space

Constraints

1 ≤ drivers.length ≤ 10⁴
1 ≤ rides.length ≤ 10⁴
1 ≤ acceptedRides.length ≤ rides.length
All dates are valid and in proper format
Each driver_id is unique in Drivers table
Each ride_id is unique in Rides and AcceptedRides tables
AcceptedRides.ride_id always exists in Rides table

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Input & Output

Visualization

Time & Space Complexity

Related Problems

Constraints

Common Approaches

Brute Force (Multiple Subqueries) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Constraints

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