Description

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.

Solutions

Solution 1: Sorted Set + Hash Table

We use two sorted sets $\textit{riders}$ and $\textit{drivers}$ to store waiting riders and available drivers respectively. Each element is a tuple $(t, \textit{id})$, representing the ID of the rider/driver and their timestamp $t$ when they joined the system. The timestamp $t$ is used to distinguish the order of arrival. Initially, $t = 0$, and each time a rider or driver is added, $t$ is incremented by $1$.

Additionally, we use a hash table $\textit{d}$ to store the mapping between each rider’s ID and their timestamp, which facilitates lookup when canceling a rider’s request.

Specifically:

When adding a rider, we add $(t, \textit{riderId})$ to $\textit{riders}$, set $\textit{d}[\textit{riderId}] = t$, and then increment $t$ by $1$.
When adding a driver, we add $(t, \textit{driverId})$ to $\textit{drivers}$ and then increment $t$ by $1$.
When matching a driver with a rider, if either $\textit{riders}$ or $\textit{drivers}$ is empty, we return $[-1, -1]$. Otherwise, we remove the elements with the smallest timestamps from both $\textit{riders}$ and $\textit{drivers}$, namely $(t_r, \textit{riderId})$ and $(t_d, \textit{driverId})$, and return $[\textit{driverId}, \textit{riderId}]$.
When canceling a rider’s request, we look up the rider’s timestamp $t$ through $\textit{d}$, and then remove $(t, \textit{riderId})$ from $\textit{riders}$.

The time complexity is $O(\log n)$ per operation, where $n$ is the current number of riders or drivers. The space complexity is $O(n)$.

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);
 */

class RideSharingSystem {
private:
    int t;
    set<pair<int, int>> riders;
    set<pair<int, int>> drivers;
    unordered_map<int, int> d;

public:
    RideSharingSystem() {
        t = 0;
    }

    void addRider(int riderId) {
        d[riderId] = t;
        riders.insert({t, riderId});
        t++;
    }

    void addDriver(int driverId) {
        drivers.insert({t, driverId});
        t++;
    }

    vector<int> matchDriverWithRider() {
        if (riders.empty() || drivers.empty()) {
            return {-1, -1};
        }
        int driverId = drivers.begin()->second;
        int riderId = riders.begin()->second;
        drivers.erase(drivers.begin());
        riders.erase(riders.begin());
        return {driverId, riderId};
    }

    void cancelRider(int riderId) {
        auto it = d.find(riderId);
        if (it != d.end()) {
            riders.erase({it->second, riderId});
        }
    }
};

/**
 * Your RideSharingSystem object will be instantiated and called as such:
 * RideSharingSystem* obj = new RideSharingSystem();
 * obj->addRider(riderId);
 * obj->addDriver(driverId);
 * vector<int> param_3 = obj->matchDriverWithRider();
 * obj->cancelRider(riderId);
 */

class RideSharingSystem:

    def __init__(self):
        self.t = 0
        self.riders = SortedList()
        self.drivers = SortedList()
        self.d = defaultdict(int)

    def addRider(self, riderId: int) -> None:
        self.d[riderId] = self.t
        self.riders.add((self.t, riderId))
        self.t += 1

    def addDriver(self, driverId: int) -> None:
        self.drivers.add((self.t, driverId))
        self.t += 1

    def matchDriverWithRider(self) -> List[int]:
        if len(self.riders) < 1 or len(self.drivers) < 1:
            return [-1, -1]
        return [self.drivers.pop(0)[1], self.riders.pop(0)[1]]

    def cancelRider(self, riderId: int) -> None:
        self.riders.discard((self.d[riderId], riderId))


# Your RideSharingSystem object will be instantiated and called as such:
# obj = RideSharingSystem()
# obj.addRider(riderId)
# obj.addDriver(driverId)
# param_3 = obj.matchDriverWithRider()
# obj.cancelRider(riderId)

type RideSharingSystem struct {
	t       int
	riders  *redblacktree.Tree[int, int]
	drivers *redblacktree.Tree[int, int]
	d       map[int]int
}

func Constructor() RideSharingSystem {
	return RideSharingSystem{
		t:       0,
		riders:  redblacktree.New[int, int](),
		drivers: redblacktree.New[int, int](),
		d:       make(map[int]int),
	}
}

func (this *RideSharingSystem) AddRider(riderId int) {
	this.d[riderId] = this.t
	this.riders.Put(this.t, riderId)
	this.t++
}

func (this *RideSharingSystem) AddDriver(driverId int) {
	this.drivers.Put(this.t, driverId)
	this.t++
}

func (this *RideSharingSystem) MatchDriverWithRider() []int {
	if this.riders.Empty() || this.drivers.Empty() {
		return []int{-1, -1}
	}

	driverTime, driverId := this.drivers.Left().Key, this.drivers.Left().Value
	riderTime, riderId := this.riders.Left().Key, this.riders.Left().Value

	this.drivers.Remove(driverTime)
	this.riders.Remove(riderTime)

	return []int{driverId, riderId}
}

func (this *RideSharingSystem) CancelRider(riderId int) {
	time, exists := this.d[riderId]
	if !exists {
		return
	}
	this.riders.Remove(time)
}

/**
 * Your RideSharingSystem object will be instantiated and called as such:
 * obj := Constructor();
 * obj.AddRider(riderId);
 * obj.AddDriver(driverId);
 * param_3 := obj.MatchDriverWithRider();
 * obj.CancelRider(riderId);
 */

3829 - Design Ride Sharing System

Description

Solutions

Solution 1: Sorted Set + Hash Table

All Problems

All Solutions

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3829. Design Ride Sharing System

Description

Solutions

Solution 1: Sorted Set + Hash Table

All Problems

All Solutions