# 4. 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
```

## Solution: 

**Approach:**\
\&#xNAN;**`To maintain a count of passenger at each time stamp`**

```cpp
class Solution
{
public:
    bool carPooling(vector<vector<int>> &trips, int capacity)
    {

        priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>> pq;

        for (int i = 0; i < trips.size(); i++)
        {

            int numPass = trips[i][0];
            int st = trips[i][1];
            int ent = trips[i][2];
            pq.push({st, numPass});
            pq.push({ent, -numPass});
        }

        int numBoarding = 0;
        while (!pq.empty())
        {
            int pass = pq.top().second;
            pq.pop();
            numBoarding += pass;
            if (numBoarding > capacity)
            {
                return false;
            }
        }

        return true;
    }
};
```

**Time Complexity: O(n logn)**


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