1094. Car Pooling

You are driving a vehicle that has capacity empty seats initially available for passengers.  The vehicle only drives east (ie. it cannot turn around and drive west.)

Given a list of trips, trip[i] = [num_passengers, start_location, end_location] contains information about the i-th trip: the number of passengers that must be picked up, and the locations to pick them up and drop them off.  The locations are given as the number of kilometers due east from your vehicle's initial location.

Return true if and only if it is possible to pick up and drop off all passengers for all the given trips. 

Example 1:

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

Example 2:

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

Example 3:

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

Example 4:

Input: trips = [[3,2,7],[3,7,9],[8,3,9]], capacity = 11
Output: true

 

Constraints:

1. trips.length <= 1000
2. trips[i].length == 3
3. 1 <= trips[i][0] <= 100
4. 0 <= trips[i][1] < trips[i][2] <= 1000
5. 1 <= capacity <= 100000

M1: greedy, sort all trips by start point, then use priorityqueue to store trips, order by end point

time = O(nlogn), space = O(n)

class Solution {
    public boolean carPooling(int[][] trips, int capacity) {
        Arrays.sort(trips, (a, b) -> a[1] - b[1]);  // sort by start location
        PriorityQueue<int[]> pq = new PriorityQueue<>((a, b) -> a[2] - b[2]);   // order by end location
        int count = capacity;
        for(int[] trip : trips) {
            while(!pq.isEmpty() && trip[1] >= pq.peek()[2]) {    // drop off
                count += pq.poll()[0];
            }
            count -= trip[0];   // pick up
            if(count < 0) {
                return false;
            }
            pq.offer(trip);
        }
        return true;
    }
}

M2: count passengers at every stop (since stops number are limited)

Process all trips, adding passenger count to the start location, and removing it from the end location.
After processing all trips, a positive value for the specific location tells that we are getting more passengers; negative - more empty seats.
Finally, scan all stops and check if we ever exceed our vehicle capacity.
time = O(n), space = O(1)

class Solution {
    public boolean carPooling(int[][] trips, int capacity) {
        int[] stops = new int[1001];
        for(int[] t : trips) {
            stops[t[1]] += t[0];
            stops[t[2]] -= t[0];
        }
        
        for(int i = 0; i < stops.length && capacity >= 0; i++) {
            capacity -= stops[i];
        }
        return capacity >= 0;
    }
}