1396. Design Underground System
Description
Implement the UndergroundSystem
class:
void checkIn(int id, string stationName, int t)
A customer with a card id equal to
id
, gets in the stationstationName
at timet
.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
, gets out from the stationstationName
at timet
.
double getAverageTime(string startStation, string endStation)
Returns the average time to travel between the
startStation
and theendStation
.The average time is computed from all the previous traveling from
startStation
toendStation
that happened directly.Call to
getAverageTime
is always valid.
You can assume all calls to checkIn
and checkOut
methods are consistent. If a customer gets in at time t1 at some station, they get out at time t2 with t2 > t1. All events happen in chronological order.
Constraints
Approach
Links
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Examples
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);
undergroundSystem.checkOut(27, "Waterloo", 20);
undergroundSystem.checkOut(32, "Cambridge", 22);
undergroundSystem.getAverageTime("Paradise", "Cambridge"); // return 14.00000. There was only one travel from "Paradise" (at time 8) to "Cambridge" (at time 22)
undergroundSystem.getAverageTime("Leyton", "Waterloo"); // return 11.00000. 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 = 11.00000
undergroundSystem.checkIn(10, "Leyton", 24);
undergroundSystem.getAverageTime("Leyton", "Waterloo"); // return 11.00000
undergroundSystem.checkOut(10, "Waterloo", 38);
undergroundSystem.getAverageTime("Leyton", "Waterloo"); // return 12.00000
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