714171. Bus Conductor
Bus Conductor
Slug: bus-conductor--170647 Difficulty: Easy Id: 714171 Topic Tags: Arrays, Sorting Company Tags: None
Summary
The Bus Conductor problem involves arranging a set of buses in a specific order based on their arrival times and departure times. The key concepts involved are arrays and sorting.
Detailed Explanation
Given an array of bus arrival and departure times, find the sequence of buses that should be arranged at each stop in such a way that no two buses arrive or depart simultaneously. This problem is equivalent to finding the maximum number of buses that can be served at a stop without any collision.
To solve this problem, we first sort the bus arrival and departure times in ascending order. Then, iterate through the sorted array, keeping track of the current bus being served (if any) and the next available time slot. If the current bus is departing or there's no bus being served, update the current bus to the one arriving at the next available time slot.
Here's a step-by-step breakdown:
- Sort the bus arrival and departure times in ascending order.
- Initialize variables:
current_bus,next_available_time, andmax_buses_served. - Iterate through the sorted array:
- If the current bus is departing or there's no bus being served, update
current_busto the one arriving atnext_available_time. - Update
next_available_timeto the departure time of the current bus.
- If the current bus is departing or there's no bus being served, update
- Return the sequence of buses that should be arranged at each stop.
Time complexity: O(n log n) due to sorting Space complexity: O(1) as we only use a constant amount of space
Optimized Solutions
Java
public class BusConductor {
public static String[] busConductor(int[][] buses) {
Arrays.sort(buses, (a, b) -> a[0] - b[0]);
int currentBus = 0;
int nextAvailableTime = buses[0][0];
List<String> result = new ArrayList<>();
for (int i = 0; i < buses.length; i++) {
if (currentBus == buses[i].length - 1 || buses[i][1] > nextAvailableTime) {
currentBus = 0;
nextAvailableTime = buses[i][0];
}
result.add("Bus " + (i + 1));
}
return result.toArray(new String[0]);
}
}