to_i respectively. The locations are given as the number of kilometers due east from the car's initial location.

Return true if it is possible to pick up and drop off all passengers for all the given trips, or false otherwise.

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

Constraints:

1 <= trips.length <= 1000
trips[i].length == 3
1 <= numPassengers_i <= 100
0 <= from_i < to_i <= 1000
1 <= capacity <= 10⁵

23:T48b,# Intuition\nSince we only have 1,001 ```stops```, we can just figure out how many people get it and out in each location. \n# Solution\nProcess 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. \n\nFinally, scan all stops and check if we ever exceed our vehicle capacity.\n## C++\n```\nbool carPooling(vector>& trips, int capacity) {\n int stops[1001] = {};\n for (auto t : trips) stops[t[1]] += t[0], stops[t[2]] -= t[0];\n for (auto i = 0; capacity >= 0 && i < 1001; ++i) capacity -= stops[i];\n return capacity >= 0;\n}\n```\n## Java\n```\npublic boolean carPooling(int[][] trips, int capacity) { \n int stops[] = new int[1001]; \n for (int t[] : trips) {\n stops[t[1]] += t[0];\n stops[t[2]] -= t[0];\n }\n for (int i = 0; capacity >= 0 && i < 1001; ++i) capacity -= stops[i];\n return capacity >= 0;\n

Car Pooling

Problem Description

Problem Description

Answer Panel