An underground railway system is keeping track of customer travel times between different stations. They are using this data to calculate the average time it takes to travel from one station to another.

Implement the UndergroundSystem class:

You may assume all calls to the checkIn and checkOut methods are consistent. If a customer checks in at time t1 then checks out at time t2, then t1 < t2. All events happen in chronological order.

Input: Operations in this order
"checkIn" : [45,"Leyton",3]
"checkIn" : [32,"Paradise",8]
"checkIn" : [27,"Leyton",10]
"checkOut" : [45,"Waterloo",15]
"checkOut" : [27,"Waterloo",20]
"checkOut" : [32,"Cambridge",22]
"getAverageTime" : ["Paradise","Cambridge"]
"getAverageTime" : ["Leyton","Waterloo"]
"checkIn" : [10,"Leyton",24]
"getAverageTime" : ["Leyton","Waterloo"]
"checkOut" : [10,"Waterloo",38]
"getAverageTime" : ["Leyton","Waterloo"]
Output:
Operation 1 "checkIn" : void
Operation 2 "checkIn" : void
Operation 3 "checkIn" : void
Operation 4 "checkOut" : void
Operation 5 "checkOut" : void
Operation 6 "checkOut" : void
Operation 7 "getAverageTime" : 14.00000
Operation 8 "getAverageTime" : 11.00000
Operation 9 "checkIn" : void
Operation 10 "getAverageTime" : 11.00000
Operation 11 "checkOut" : void
Operation 12 "getAverageTime" : 12.00000
Explanation:
undergroundSystem.checkIn(45, "Leyton", 3);
undergroundSystem.checkIn(32, "Paradise", 8);
undergroundSystem.checkIn(27, "Leyton", 10);
undergroundSystem.checkOut(45, "Waterloo", 15); // Customer 45 "Leyton" -> "Waterloo" in 15-3 = 12
undergroundSystem.checkOut(27, "Waterloo", 20); // Customer 27 "Leyton" -> "Waterloo" in 20-10 = 10
undergroundSystem.checkOut(32, "Cambridge", 22); // Customer 32 "Paradise" -> "Cambridge" in 22-8 = 14
undergroundSystem.getAverageTime("Paradise", "Cambridge"); // return 14.00000. One trip "Paradise" -> "Cambridge", (14) / 1 = 14
undergroundSystem.getAverageTime("Leyton", "Waterloo"); // return 11.00000. Two trips "Leyton" -> "Waterloo", (10 + 12) / 2 = 11
undergroundSystem.checkIn(10, "Leyton", 24);
undergroundSystem.getAverageTime("Leyton", "Waterloo"); // return 11.00000
undergroundSystem.checkOut(10, "Waterloo", 38); // Customer 10 "Leyton" -> "Waterloo" in 38-24 = 14
undergroundSystem.getAverageTime("Leyton", "Waterloo"); // return 12.00000. Three trips "Leyton" -> "Waterloo", (10 + 12 + 14) / 3 = 12
Constraints:

Contents

In this approach, we will use two HashMap's to store checkin and checkout time information for calculating the average.

Implementation steps:
import java.util.AbstractMap.SimpleEntry; import java.util.HashMap; public class UndergroundSystem { HashMap<Integer, SimpleEntry<String, Integer>> checkInMap = new HashMap<>(); // Id -> (stationName, t) HashMap<SimpleEntry<String, String>, int[]> totalTimeToCountMap = new HashMap<>(); //(startStation, endStation) // -> (totalTime, count) public UndergroundSystem() {} public void checkIn(int id, String stationName, int t) { checkInMap.put(id, new SimpleEntry(stationName, t)); } public void checkOut(int id, String stationName, int t) { SimpleEntry<String, Integer> checkInTime = checkInMap.get(id); SimpleEntry<String, String> startEndStationsKey = new SimpleEntry<>(checkInTime.getKey(), stationName); int[] timesCount = new int[]{t - checkInTime.getValue(), 1}; totalTimeToCountMap.merge(startEndStationsKey, timesCount, (oldVal, newVal) -> { oldVal[0]+= newVal[0]; oldVal[1]+= newVal[1]; return oldVal; }); } public double getAverageTime(String startStation, String endStation) { SimpleEntry<String, String> key = new SimpleEntry<>(startStation, endStation); int[] countTimes = totalTimeToCountMap.get(key); return (double)countTimes[0]/countTimes[1]; } public static void main(String[] args) { UndergroundSystem undergroundSystem = new UndergroundSystem(); undergroundSystem.checkIn(45, "Leyton", 3); undergroundSystem.checkIn(32, "Paradise", 8); undergroundSystem.checkIn(27, "Leyton", 10); undergroundSystem.checkOut(45, "Waterloo", 15); undergroundSystem.checkOut(27, "Waterloo", 20); undergroundSystem.checkOut(32, "Cambridge", 22); System.out.println(undergroundSystem.getAverageTime("Paradise", "Cambridge")); System.out.println(undergroundSystem.getAverageTime("Leyton", "Waterloo")); undergroundSystem.checkIn(10, "Leyton", 24); System.out.println(undergroundSystem.getAverageTime("Leyton", "Waterloo")); undergroundSystem.checkOut(10, "Waterloo", 38); System.out.println(undergroundSystem.getAverageTime("Leyton", "Waterloo")); } }
Complexity Analysis:

Time complexity: Above code runs in O(1) since we are using HashMaps and that returns in constant time.
Space complexity: O(n), since we are storing information into HashMap, where n is total number of unique travels made, that is between a (startStation, endStation).

Above implementations source code can be found at GitHub link for Java code