How do you solve Design Ride Sharing System on LeetCode?

Design Ride Sharing System (LeetCode #3829) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using queues preserves order for matching.

What is the time complexity of Design Ride Sharing System?

The optimal solution for Design Ride Sharing System runs in O(n) time and uses O(n) space. The complexity is linear due to efficient queue operations for adding, matching, and canceling riders and drivers.

What is the brute force solution for Design Ride Sharing System?

The brute force approach for Design Ride Sharing System is Brute Force, with O(n²) time complexity and O(1) space complexity. This approach checks each rider against each driver to find a match. It’s straightforward but inefficient due to repeated checks. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Design Ride Sharing System?

Design Ride Sharing System uses the following concepts: Hash Table, Design, Queue, Data Stream. The recommended patterns to study are: Queue, Data Stream.

What are the interview tips for Design Ride Sharing System?

Explain your thought process clearly. Discuss trade-offs of different data structures. Practice coding on a whiteboard or shared screen.

What are common mistakes when solving Design Ride Sharing System?

Not handling cancellations correctly. Forgetting to check if queues are empty before matching.

#3829

Design Ride Sharing System

Medium

Hash TableDesignQueueData StreamQueueData Stream

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

Using queues allows us to efficiently manage the order of riders and drivers, ensuring quick matching and cancellations.

⚙️

Algorithm

3 steps

1Step 1: Use two queues to store riders and drivers.
2Step 2: When matching, dequeue the first rider and driver if both are available.
3Step 3: For cancellation, remove the rider from the queue efficiently.

solution.py15 lines

1from collections import deque
2class RideSharingSystem:
3    def __init__(self):
4        self.riders = deque()
5        self.drivers = deque()
6    def addRider(self, riderId):
7        self.riders.append(riderId)
8    def addDriver(self, driverId):
9        self.drivers.append(driverId)
10    def matchDriverWithRider(self):
11        if self.riders and self.drivers:
12            return [self.drivers.popleft(), self.riders.popleft()]
13        return [-1, -1]
14    def cancelRider(self, riderId):
15        self.riders = deque([r for r in self.riders if r != riderId])

ℹ

Complexity note: The complexity is linear due to efficient queue operations for adding, matching, and canceling riders and drivers.

1Using queues preserves order for matching.
2Efficient cancellation requires a structure that supports quick removal.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.