LeetCode #3829 — MEDIUM

Design Ride Sharing System

Move from brute-force thinking to an efficient approach using hash map strategy.

Solve on LeetCode
The Problem

Problem Statement

A ride sharing system manages ride requests from riders and availability from drivers. Riders request rides, and drivers become available over time. The system should match riders and drivers in the order they arrive.

Implement the RideSharingSystem class:

  • RideSharingSystem() Initializes the system.
  • void addRider(int riderId) Adds a new rider with the given riderId.
  • void addDriver(int driverId) Adds a new driver with the given driverId.
  • int[] matchDriverWithRider() Matches the earliest available driver with the earliest waiting rider and removes both of them from the system. Returns an integer array of size 2 where result = [driverId, riderId] if a match is made. If no match is available, returns [-1, -1].
  • void cancelRider(int riderId) Cancels the ride request of the rider with the given riderId if the rider exists and has not yet been matched.

Example 1:

Input:
["RideSharingSystem", "addRider", "addDriver", "addRider", "matchDriverWithRider", "addDriver", "cancelRider", "matchDriverWithRider", "matchDriverWithRider"]
[[], [3], [2], [1], [], [5], [3], [], []]

Output:
[null, null, null, null, [2, 3], null, null, [5, 1], [-1, -1]]

Explanation

RideSharingSystem rideSharingSystem = new RideSharingSystem(); // Initializes the system
rideSharingSystem.addRider(3); // rider 3 joins the queue
rideSharingSystem.addDriver(2); // driver 2 joins the queue
rideSharingSystem.addRider(1); // rider 1 joins the queue
rideSharingSystem.matchDriverWithRider(); // returns [2, 3]
rideSharingSystem.addDriver(5); // driver 5 becomes available
rideSharingSystem.cancelRider(3); // rider 3 is already matched, cancel has no effect
rideSharingSystem.matchDriverWithRider(); // returns [5, 1]
rideSharingSystem.matchDriverWithRider(); // returns [-1, -1]

Example 2:

Input:
["RideSharingSystem", "addRider", "addDriver", "addDriver", "matchDriverWithRider", "addRider", "cancelRider", "matchDriverWithRider"]
[[], [8], [8], [6], [], [2], [2], []]

Output:
[null, null, null, null, [8, 8], null, null, [-1, -1]]

Explanation

RideSharingSystem rideSharingSystem = new RideSharingSystem(); // Initializes the system
rideSharingSystem.addRider(8); // rider 8 joins the queue
rideSharingSystem.addDriver(8); // driver 8 joins the queue
rideSharingSystem.addDriver(6); // driver 6 joins the queue
rideSharingSystem.matchDriverWithRider(); // returns [8, 8]
rideSharingSystem.addRider(2); // rider 2 joins the queue
rideSharingSystem.cancelRider(2); // rider 2 cancels
rideSharingSystem.matchDriverWithRider(); // returns [-1, -1]

Constraints:

  • 1 <= riderId, driverId <= 1000
  • Each riderId is unique among riders and is added at most once.
  • Each driverId is unique among drivers and is added at most once.
  • At most 1000 calls will be made in total to addRider​​​​​​​, addDriver, matchDriverWithRider, and cancelRider.
Patterns Used
🏗Design Patterns→

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: A ride sharing system manages ride requests from riders and availability from drivers. Riders request rides, and drivers become available over time. The system should match riders and drivers in the order they arrive. Implement the RideSharingSystem class: RideSharingSystem() Initializes the system. void addRider(int riderId) Adds a new rider with the given riderId. void addDriver(int driverId) Adds a new driver with the given driverId. int[] matchDriverWithRider() Matches the earliest available driver with the earliest waiting rider and removes both of them from the system. Returns an integer array of size 2 where result = [driverId, riderId] if a match is made. If no match is available, returns [-1, -1]. void cancelRider(int riderId) Cancels the ride request of the rider with the given riderId if the rider exists and has not yet been matched.

Baseline thinking

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

Pattern signal: Hash Map · Design

Example 1

["RideSharingSystem","addRider","addDriver","addRider","matchDriverWithRider","addDriver","cancelRider","matchDriverWithRider","matchDriverWithRider"]
[[],[3],[2],[1],[],[5],[3],[],[]]

Example 2

["RideSharingSystem","addRider","addDriver","addDriver","matchDriverWithRider","addRider","cancelRider","matchDriverWithRider"]
[[],[8],[8],[6],[],[2],[2],[]]
Step 02

Core Insight

What unlocks the optimal approach

  • Use queues to preserve FIFO order for riders and drivers.
  • Track waiting riders with a structure that supports efficient removal on cancel (for example, a list plus a map).
  • Before matching, skip or remove canceled riders to ensure the earliest valid rider is paired with the earliest driver.
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. 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 #3829: Design Ride Sharing System
class RideSharingSystem {
    private int t;
    private TreeSet<int[]> riders;
    private TreeSet<int[]> drivers;
    private Map<Integer, Integer> d;

    public RideSharingSystem() {
        this.t = 0;
        this.riders = new TreeSet<>(
            (a, b) -> a[0] != b[0] ? Integer.compare(a[0], b[0]) : Integer.compare(a[1], b[1]));
        this.drivers = new TreeSet<>(
            (a, b) -> a[0] != b[0] ? Integer.compare(a[0], b[0]) : Integer.compare(a[1], b[1]));
        this.d = new HashMap<>();
    }

    public void addRider(int riderId) {
        d.put(riderId, t);
        riders.add(new int[] {t, riderId});
        t++;
    }

    public void addDriver(int driverId) {
        drivers.add(new int[] {t, driverId});
        t++;
    }

    public int[] matchDriverWithRider() {
        if (riders.isEmpty() || drivers.isEmpty()) {
            return new int[] {-1, -1};
        }
        int driverId = drivers.pollFirst()[1];
        int riderId = riders.pollFirst()[1];
        return new int[] {driverId, riderId};
    }

    public void cancelRider(int riderId) {
        Integer time = d.get(riderId);
        if (time != null) {
            riders.remove(new int[] {time, riderId});
        }
    }
}

/**
 * Your RideSharingSystem object will be instantiated and called as such:
 * RideSharingSystem obj = new RideSharingSystem();
 * obj.addRider(riderId);
 * obj.addDriver(driverId);
 * int[] param_3 = obj.matchDriverWithRider();
 * obj.cancelRider(riderId);
 */
Step 06

Interactive Study Demo

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

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(1) per op
Space
O(n)

Approach Breakdown

NAIVE
O(n) per op time
O(n) space

Use a simple list or array for storage. Each operation (get, put, remove) requires a linear scan to find the target element — O(n) per operation. Space is O(n) to store the data. The linear search makes this impractical for frequent operations.

OPTIMIZED DESIGN
O(1) per op time
O(n) space

Design problems target O(1) amortized per operation by combining data structures (hash map + doubly-linked list for LRU, stack + min-tracking for MinStack). Space is always at least O(n) to store the data. The challenge is achieving constant-time operations through clever structure composition.

Shortcut: Combine two data structures to get O(1) for each operation type. Space is always O(n).
Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

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