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refactor 1396
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src/main/java/com/fishercoder/solutions/_1396.java

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import java.util.LinkedList;
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import java.util.Map;
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/**
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* 1396. Design Underground System
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*
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* Implement the class UndergroundSystem that supports three methods:
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*
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* 1. checkIn(int id, string stationName, int t)
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* A customer with id card equal to id, gets in the station stationName at time t.
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* A customer can only be checked into one place at a time.
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* 2. checkOut(int id, string stationName, int t)
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* A customer with id card equal to id, gets out from the station stationName at time t.
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* 3. getAverageTime(string startStation, string endStation)
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* Returns the average time to travel between the startStation and the endStation.
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* The average time is computed from all the previous traveling from startStation to endStation that happened directly.
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* Call to getAverageTime is always valid.
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* You can assume all calls to checkIn and checkOut methods are consistent. That is, if a customer gets in at time t1 at some station, then it gets out at time t2 with t2 > t1. All events happen in chronological order.
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*
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* Example 1:
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* Input
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* ["UndergroundSystem","checkIn","checkIn","checkIn","checkOut","checkOut","checkOut","getAverageTime","getAverageTime","checkIn","getAverageTime","checkOut","getAverageTime"]
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* [[],[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"]]
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* Output
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* [null,null,null,null,null,null,null,14.0,11.0,null,11.0,null,12.0]
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*
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* Explanation
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* UndergroundSystem undergroundSystem = new UndergroundSystem();
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* undergroundSystem.checkIn(45, "Leyton", 3);
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* undergroundSystem.checkIn(32, "Paradise", 8);
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* undergroundSystem.checkIn(27, "Leyton", 10);
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* undergroundSystem.checkOut(45, "Waterloo", 15);
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* undergroundSystem.checkOut(27, "Waterloo", 20);
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* undergroundSystem.checkOut(32, "Cambridge", 22);
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* undergroundSystem.getAverageTime("Paradise", "Cambridge"); // return 14.0. There was only one travel from "Paradise" (at time 8) to "Cambridge" (at time 22)
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* undergroundSystem.getAverageTime("Leyton", "Waterloo"); // return 11.0. There were two travels from "Leyton" to "Waterloo", a customer with id=45 from time=3 to time=15 and a customer with id=27 from time=10 to time=20. So the average time is ( (15-3) + (20-10) ) / 2 = 11.0
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* undergroundSystem.checkIn(10, "Leyton", 24);
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* undergroundSystem.getAverageTime("Leyton", "Waterloo"); // return 11.0
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* undergroundSystem.checkOut(10, "Waterloo", 38);
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* undergroundSystem.getAverageTime("Leyton", "Waterloo"); // return 12.0
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*
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* Constraints:
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* There will be at most 20000 operations.
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* 1 <= id, t <= 10^6
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* All strings consist of uppercase, lowercase English letters and digits.
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* 1 <= stationName.length <= 10
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* Answers within 10^-5 of the actual value will be accepted as correct.
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* */
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public class _1396 {
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public static class Solution1 {
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public class UndergroundSystem {

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