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

Solution

Time complexity : O(n)
Space complexity : O(1001)

class Solution {
public:
    bool carPooling(vector<vector<int>>& trips, int capacity) {
        vector<int> delta(1001, 0);
        
        for (const auto& trip: trips) {
            delta[trip[1]] -= trip[0];
            delta[trip[2]] += trip[0];
        }
        
        for (const int d: delta)
            if ((capacity += d) < 0)
                return false;
        
        return true;
    }
};

記錄每一站(最多1000站)，的乘客變化，即該站總共上車多少人、下車多少人。

從頭，用初始容量，一站一站去接受載客量變化，若途中容量為負，則返回false。

若一切順利，返回true。