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Car Pooling⚓︎

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Description⚓︎

There is a car with capacity empty seats. The vehicle only drives east (i.e., it cannot turn around and drive west).

You are given the integer capacity and an array trips where trips[i] = [numPassengers_i, from_i, to_i] indicates that the i^th trip has numPassengersi passengers and the locations to pick them up and drop them off are from_i and 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:

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

Example 2:

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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 <= 10^5

Solution⚓︎

See reference (Chinese).

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class Solution {
public:
    bool carPooling(vector<vector<int>>& trips, int capacity) {
        int n = trips.size();
        vector<int> count(1001);
        for (auto& trip : trips) {
            int num = trip[0], from = trip[1], to = trip[2];
            count[from] += num;
            count[to] -= num;
        }

        int sum = 0;
        for (int element : count) {
            sum += element;
            if (sum > capacity) return false;
        }

        return true;
    }
};
  • Time complexity: \(O(n+U)\), where \(n\) is the length of trips and \(U=\max (to_i)\);
  • Space complexity: \(O(U)\).