Design Underground System - LeetCode solutions

Problem

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:

void checkIn(int id, string stationName, int t)

A customer with a card ID equal to id, checks in at the station stationName at time t.
A customer can only be checked into one place at a time.

void checkOut(int id, string stationName, int t)
A customer with a card ID equal to id, checks out from the station stationName at time t.

double getAverageTime(string startStation, string endStation)
Returns the average time it takes to travel from startStation to endStation.
The average time is computed from all the previous traveling times from startStation to endStation that happened directly, meaning a check in at startStation followed by a check out from endStation.
The time it takes to travel from startStation to endStation may be different from the time it takes to travel from endStation to startStation.
There will be at least one customer that has traveled from startStation to endStation before getAverageTime is called.

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.

Example 1:

Input
["UndergroundSystem","checkIn","checkIn","checkIn","checkOut","checkOut","checkOut","getAverageTime","getAverageTime","checkIn","getAverageTime","checkOut","getAverageTime"]
[[],[45,"Leyton",3],[32,"Paradise",8],[27,"Leyton",10],[45,"Waterloo",15],[27,"Waterloo",20],[32,"Cambridge",22],["Paradise","Cambridge"],["Leyton","Waterloo"],[10,"Leyton",24],["Leyton","Waterloo"],[10,"Waterloo",38],["Leyton","Waterloo"]]

Output
[null,null,null,null,null,null,null,14.00000,11.00000,null,11.00000,null,12.00000]

Explanation
UndergroundSystem undergroundSystem = new UndergroundSystem();
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

Example 2:

Input
["UndergroundSystem","checkIn","checkOut","getAverageTime","checkIn","checkOut","getAverageTime","checkIn","checkOut","getAverageTime"]
[[],[10,"Leyton",3],[10,"Paradise",8],["Leyton","Paradise"],[5,"Leyton",10],[5,"Paradise",16],["Leyton","Paradise"],[2,"Leyton",21],[2,"Paradise",30],["Leyton","Paradise"]]

Output
[null,null,null,5.00000,null,null,5.50000,null,null,6.66667]

Explanation
UndergroundSystem undergroundSystem = new UndergroundSystem();
undergroundSystem.checkIn(10, "Leyton", 3);
undergroundSystem.checkOut(10, "Paradise", 8); // Customer 10 "Leyton" -> "Paradise" in 8-3 = 5
undergroundSystem.getAverageTime("Leyton", "Paradise"); // return 5.00000, (5) / 1 = 5
undergroundSystem.checkIn(5, "Leyton", 10);
undergroundSystem.checkOut(5, "Paradise", 16); // Customer 5 "Leyton" -> "Paradise" in 16-10 = 6
undergroundSystem.getAverageTime("Leyton", "Paradise"); // return 5.50000, (5 + 6) / 2 = 5.5
undergroundSystem.checkIn(2, "Leyton", 21);
undergroundSystem.checkOut(2, "Paradise", 30); // Customer 2 "Leyton" -> "Paradise" in 30-21 = 9
undergroundSystem.getAverageTime("Leyton", "Paradise"); // return 6.66667, (5 + 6 + 9) / 3 = 6.66667

Constraints:

1 <= id, t <= 10^6
1 <= stationName.length, startStation.length, endStation.length <= 10
All strings consist of uppercase and lowercase English letters and digits.
There will be at most 2 * 10^4 calls in total to checkIn, checkOut, and getAverageTime.
Answers within 10-5 of the actual value will be accepted.

Solution (Java)

class UndergroundSystem {
    private static class StationAndTime {
        String stationName;
        int t;

        public StationAndTime(String stationName, int t) {
            this.stationName = stationName;
            this.t = t;
        }

        public String getStation() {
            return this.stationName;
        }

        public int getTime() {
            return this.t;
        }
    }

    private Map<String, double[]> averageTimeMap;
    private Map<Integer, LinkedList<StationAndTime>> travelerMap;

    public UndergroundSystem() {
        averageTimeMap = new HashMap<>();
        travelerMap = new HashMap<>();
    }

    public void checkIn(int id, String stationName, int t) {
        travelerMap.putIfAbsent(id, new LinkedList<>());
        travelerMap.get(id).add(new StationAndTime(stationName, t));
    }

    public void checkOut(int id, String stationName, int t) {
        LinkedList<StationAndTime> list = travelerMap.get(id);
        StationAndTime stationAndTime = list.getLast();
        String startToEndStation = stationAndTime.getStation() + "->" + stationName;
        int duration = t - stationAndTime.getTime();
        if (averageTimeMap.containsKey(startToEndStation)) {
            double[] pair = averageTimeMap.get(startToEndStation);
            double newAverage = (pair[0] * pair[1] + duration) / (pair[1] + 1);
            averageTimeMap.put(startToEndStation, new double[] {newAverage, pair[1] + 1});
        } else {
            averageTimeMap.put(startToEndStation, new double[] {duration, 1});
        }
    }

    public double getAverageTime(String startStation, String endStation) {
        return averageTimeMap.get(startStation + "->" + endStation)[0];
    }
}

/**
 * Your UndergroundSystem object will be instantiated and called as such:
 * UndergroundSystem obj = new UndergroundSystem();
 * obj.checkIn(id,stationName,t);
 * obj.checkOut(id,stationName,t);
 * double param_3 = obj.getAverageTime(startStation,endStation);
 */

Explain:

nope.

Complexity:

Time complexity : O(n).
Space complexity : O(n).