How do you solve Car Pooling on LeetCode?

Car Pooling (LeetCode #1094) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Using a sweep line technique can significantly reduce complexity.

What is the time complexity of Car Pooling?

The optimal solution for Car Pooling runs in O(n log n) time and uses O(n) space. The time complexity is O(n log n) due to the sorting of events, while the space complexity is O(n) for storing the events.

What is the brute force solution for Car Pooling?

The brute force approach for Car Pooling is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute force approach, we will simulate each trip and keep track of the number of passengers in the car at each point. We check at every pickup and drop-off if the number of passengers exceeds the car's capacity. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Car Pooling?

Car Pooling uses the following concepts: Array, Sorting, Heap (Priority Queue), Simulation, Prefix Sum. The recommended patterns to study are: Sweep Line, Sorting, Array.

What are the interview tips for Car Pooling?

Always clarify constraints and edge cases with the interviewer. Explain your thought process while coding to demonstrate understanding. Practice similar problems to recognize patterns and improve problem-solving speed.

What are common mistakes when solving Car Pooling?

Not considering the order of events (pickup before drop-off). Failing to handle edge cases where capacity is exactly met.

#1094

Car Pooling

Medium

ArraySortingHeap (Priority Queue)SimulationPrefix SumSweep LineSortingArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n log n)Space O(n)

The optimal solution uses a sweep line technique where we treat each trip as two events: a pickup and a drop-off. By sorting these events, we can efficiently track the number of passengers in the car at any point.

⚙️

Algorithm

3 steps

1Step 1: Create an array of events where each trip generates two events: one for pickup and one for drop-off.
2Step 2: Sort the events first by location, then by type (pickup before drop-off).
3Step 3: Traverse through the sorted events, adjusting the current passenger count and checking against the capacity.

solution.py14 lines

1# Full working Python code
2
3def carPooling(trips, capacity):
4    events = []
5    for num_passengers, from_loc, to_loc in trips:
6        events.append((from_loc, num_passengers))  # pickup
7        events.append((to_loc, -num_passengers))  # drop-off
8    events.sort()
9    current_passengers = 0
10    for _, change in events:
11        current_passengers += change
12        if current_passengers > capacity:
13            return False
14    return True

ℹ

Complexity note: The time complexity is O(n log n) due to the sorting of events, while the space complexity is O(n) for storing the events.

1Using a sweep line technique can significantly reduce complexity.
2Sorting events allows us to process pickups and drop-offs in a single pass.

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