How do I practice coding and algorithm questions?

Use PracHub's coding console to write, test, and debug your solutions in Python or JavaScript. View hints, test against sample inputs, and compare with official solutions.

What difficulty level is this coding question?

This is a medium difficulty Coding & Algorithms question, commonly asked during Technical Screen rounds at Rippling.

What role is this question designed for?

This question is commonly asked for Software Engineer candidates at Rippling during technical interviews.

Design a Driver Payroll System | Rippling Coding Question

Quick Overview

This question evaluates a candidate's understanding of data structure design, time-series aggregation, and algorithmic complexity for in-memory accounting of per-driver earnings.

Design a Driver Payroll System

Company: Rippling

Role: Software Engineer

Category: Coding & Algorithms

Difficulty: medium

Interview Round: Technical Screen

Design an in-memory payroll system for a food delivery company. Each driver has a unique `driver_id` and a fixed `hourly_rate` set when the driver is registered. Implement a data structure that supports the following operations efficiently: 1. `register_driver(driver_id, hourly_rate)` - Add a new driver to the system. 2. `record_shift(driver_id, start_ts, end_ts)` - Record one completed shift for the driver. - `start_ts` and `end_ts` are Unix timestamps in seconds, and `end_ts > start_ts`. - The pay for the shift is `hourly_rate * (end_ts - start_ts) / 3600`. 3. `get_total_earned(driver_id)` - Return the total earnings accumulated by that driver across all recorded shifts. Follow-up: 4. `settle_up_to(cutoff_ts)` - Pay out all earnings from shifts that ended at or before `cutoff_ts`. 5. `get_unpaid_balance(driver_id)` - After one or more settlements, return how much money is still unpaid for that driver. Discuss what data structures you would use, how each operation works, and the time and space complexity of your design. If a driver has many historical shifts, explain how to support the settlement operation efficiently.

Quick Answer: This question evaluates a candidate's understanding of data structure design, time-series aggregation, and algorithmic complexity for in-memory accounting of per-driver earnings.

Part 1: Basic Driver Payroll Tracker

Implement a basic in-memory payroll tracker for delivery drivers. Each driver has a unique string `driver_id` and a fixed `hourly_rate` set at registration time. You are given a list of operations to process in order. Supported operations: - `('register_driver', driver_id, hourly_rate)` — add a new driver. - `('record_shift', driver_id, start_ts, end_ts)` — record one completed shift. The pay for the shift is `hourly_rate * (end_ts - start_ts) / 3600`. - `('get_total_earned', driver_id)` — query the total amount earned by that driver across all recorded shifts so far. Return the results of all `get_total_earned` operations in the order they appear. All operations are valid: - a driver is registered before any shift/query for that driver, - each `driver_id` is registered once, - `end_ts > start_ts`.

Constraints

0 <= len(operations) <= 2 * 10^5
1 <= hourly_rate <= 10^6
0 <= start_ts < end_ts <= 10^9
Each driver_id is a non-empty string and is registered exactly once
All query and shift operations refer to already-registered drivers

Examples

Loading coding console...

Part 2: Driver Payroll with Settlement Cutoffs

Extend the payroll system to support settlements. Each driver has a unique string `driver_id` and a fixed `hourly_rate`. You are given a list of operations to process in order. Supported operations: - `('register_driver', driver_id, hourly_rate)` — add a new driver. - `('record_shift', driver_id, start_ts, end_ts)` — record one completed shift. The shift's pay is `hourly_rate * (end_ts - start_ts) / 3600`. - `('settle_up_to', cutoff_ts)` — mark as paid every shift whose `end_ts <= cutoff_ts`. - `('get_unpaid_balance', driver_id)` — return how much money is still unpaid for that driver. Return the results of all `get_unpaid_balance` operations in the order they appear. Your implementation should support settlement efficiently even when there are many historical shifts. Avoid scanning all recorded shifts on every settlement. All operations are valid: - a driver is registered before any shift/query for that driver, - each `driver_id` is registered once, - `end_ts > start_ts`.

Constraints

0 <= len(operations) <= 2 * 10^5
1 <= hourly_rate <= 10^6
0 <= start_ts < end_ts <= 10^9
Each driver_id is a non-empty string and is registered exactly once
All query and shift operations refer to already-registered drivers

Examples

Input: [('register_driver', 'd1', 20), ('record_shift', 'd1', 0, 3600), ('record_shift', 'd1', 3600, 10800), ('get_unpaid_balance', 'd1'), ('settle_up_to', 4000), ('get_unpaid_balance', 'd1'), ('settle_up_to', 20000), ('get_unpaid_balance', 'd1')]

Expected Output: [60.0, 40.0, 0.0]

Explanation: Before any settlement, both shifts are unpaid for a total of 60. After settling up to 4000, only the first shift is paid. After settling up to 20000, both are paid.

Input: [('register_driver', 'd1', 10), ('register_driver', 'd2', 30), ('record_shift', 'd1', 0, 7200), ('record_shift', 'd2', 0, 3600), ('settle_up_to', 5000), ('get_unpaid_balance', 'd1'), ('get_unpaid_balance', 'd2'), ('record_shift', 'd2', 3600, 7200), ('get_unpaid_balance', 'd2'), ('settle_up_to', 8000), ('get_unpaid_balance', 'd1'), ('get_unpaid_balance', 'd2')]

Expected Output: [20.0, 0.0, 30.0, 0.0, 0.0]

Explanation: The first settlement pays only d2's first shift. Later, after another d2 shift is recorded and settlement reaches 8000, all remaining unpaid shifts are paid.

Input: [('register_driver', 'empty', 25), ('settle_up_to', 100000), ('get_unpaid_balance', 'empty')]

Expected Output: [0.0]

Explanation: Edge case: a driver with no shifts has unpaid balance 0, even after settlement.

Input: [('register_driver', 'x', 10), ('record_shift', 'x', 0, 3600), ('settle_up_to', 5000), ('get_unpaid_balance', 'x'), ('settle_up_to', 1000), ('get_unpaid_balance', 'x')]

Expected Output: [0.0, 0.0]

Explanation: A shift should not be paid twice. A later settlement with a smaller cutoff changes nothing.

Hints

When a shift is recorded, compute its pay immediately and keep track of it as unpaid for that driver.
A min-heap ordered by shift end time lets `settle_up_to` remove only the shifts that became payable, instead of scanning every stored shift.