What is the brute force approach and complexity for Car Pooling?

For each location, sum the passengers of all trips covering it with time complexity O(N · R) and space complexity O(1).

What is the optimal approach and complexity for Car Pooling?

+p at from, −p at to in an array of size 1001; prefix-sum it and check capacity with optimal time complexity O(N + R), R = 1001 and space complexity O(R).

How do you identify the pattern for Car Pooling?

Look for key signals in the prompt: "get in at from, get out at to" indicates Each trip is an interval [from, to) carrying a weight p.; "never more than capacity at once" indicates Max over time of the active total → sweep, like Meeting Rooms II but adding p instead of 1.; "0 ≤ from < to ≤ 1000" indicates Small coordinates → a difference array replaces sorting..

Car Pooling

Medium·LeetCode #1094 Pro OnlyInterview-Level

LeetCode

Problem Statement

A car drives in one direction along a road, starting at location 0. It has capacity empty seats and never turns back.

Each trip trips[i] = [numPassengers, from, to] means numPassengers people get in when the car reaches location from and get out when it reaches location to.

Return true if the car can do every trip without ever carrying more than capacity passengers at once, and false otherwise.

People getting out at a location leave before new people get in there. So passengers dropped at 5 never sit next to passengers picked up at 5.

Input

A list trips of [numPassengers, from, to] triples, and an integer capacity.

Output

true if all trips fit, false if the car is ever over capacity.

Example 1

Input: trips = [[2,1,5],[3,3,7]], capacity = 4

Output: false

Between locations 3 and 5, both groups are in the car: 2 + 3 = 5 people, but only 4 seats.

Example 2

Input: trips = [[2,1,5],[3,3,7]], capacity = 5

Output: true

The busiest stretch has 5 people, which fits exactly.

Example 3

Input: trips = [[2,1,5],[3,5,7]], capacity = 3

Output: true

The first group gets out at 5 before the second group gets in at 5, so there are never more than 3 people in the car.

Constraints

• 1 ≤ trips.length ≤ 1000
• 1 ≤ numPassengers ≤ 100
• 0 ≤ from < to ≤ 1000
• 1 ≤ capacity ≤ 10^5

Pattern Recognition Clues

Recognising these key signals in the problem description immediately identifies the pattern:

“get in at from, get out at to”

→ Each trip is an interval [from, to) carrying a weight p.

“never more than capacity at once”

→ Max over time of the active total → sweep, like Meeting Rooms II but adding p instead of 1.

“0 ≤ from < to ≤ 1000”

→ Small coordinates → a difference array replaces sorting.

Approach Complexity Ladder

Brute Force — add up every trip at every locationO(N · R)O(1)

For each location, sum the passengers of all trips covering it

Better — sort pick-up and drop-off eventsO(N log N)O(N)

Turn each trip into +p and −p events, sort by location, sweep a running load

Optimal — difference array over locationsO(N + R), R = 1001O(R)

+p at from, −p at to in an array of size 1001; prefix-sum it and check capacity

Brute Force Intuition

For every location x from 0 to 999, look at the stretch [x, x+1) of road. Add up p for every trip with from ≤ x < to. If any sum is more than capacity, return false.

Recognition rule for this sub-pattern:

How many things are active at the same time? Sweep events in time order, or keep a min-heap of end times.

Remember this · Intervals

Trigger

Input is [start, end] pairs; overlap, merge, insert, meeting rooms, free time, minimum removals.

Think

After sorting, do I only ever need to compare against the most recent interval?

Key question

Sort by start or by end — which one makes the greedy choice obvious?

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