LeetCode 2238: Number of Times a Driver Was a Passenger

Explanation

To solve this problem, we can iterate through the list of trips and maintain a map to keep track of the number of times each driver was a passenger. For each trip, we update the count for the driver and the passenger accordingly. Finally, we return the count for the given driver.

1. Initialize a hashmap to store the count of each driver as a passenger.
2. Iterate through the list of trips:
   • For each trip, update the count of the driver and passenger.
3. Return the count for the given driver.

Time Complexity

The time complexity of this solution is O(N) where N is the number of trips.

Space Complexity

The space complexity of this solution is O(N) to store the count of each driver.

import java.util.HashMap;
import java.util.List;
import java.util.Map;

class Solution {
    public int numOfTimesDriverWasPassenger(int driver, List<List<Integer>> trips) {
        Map<Integer, Integer> passengerCount = new HashMap<>();
        
        for (List<Integer> trip : trips) {
            int driverTrip = trip.get(0);
            int passengerTrip = trip.get(1);
            
            passengerCount.put(driverTrip, passengerCount.getOrDefault(driverTrip, 0) + 1);
            passengerCount.put(passengerTrip, passengerCount.getOrDefault(passengerTrip, 0) - 1);
        }
        
        return passengerCount.getOrDefault(driver, 0);
    }
}