1094. Car Pooling.cpp

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:

trips.length <= 1000
trips[i].length == 3
1 <= trips[i][0] <= 100
0 <= trips[i][1] < trips[i][2] <= 1000
1 <= capacity <= 100000
   Hide Hint #1
Sort the pickup and dropoff events by location, then process them in order.












class Solution {
public:
    bool carPooling(vector<vector<int>>& trips, int capacity) {
        vector<int> stops (10001, 0);

        for(auto &t: trips){
            stops[t[1]]+=t[0];
            stops[t[2]]-=t[0];
        }

        int occupied=0;

        for(int i=0;capacity>=0 && i<1001;i++)
            capacity-=stops[i];

        return capacity>=0;
    }
};






/*
We only have 1001 stops
we can just figure out how many people get it and out in each location.

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.

Source:   https://leetcode.com/problems/car-pooling/discuss/317611/C%2B%2BJava-O(n)-Thousand-and-One-Stops
*/