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 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
, checks out from the stationstationName
at timet
. double getAverageTime(string startStation, string endStation)
- Returns the average time it takes to travel from
startStation
toendStation
. - The average time is computed from all the previous traveling times from
startStation
toendStation
that happened directly, meaning a check in atstartStation
followed by a check out fromendStation
. - The time it takes to travel from
startStation
toendStation
may be different from the time it takes to travel fromendStation
tostartStation
. - There will be at least one customer that has traveled from
startStation
toendStation
beforegetAverageTime
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 <= 106
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 * 104
calls in total tocheckIn
,checkOut
, andgetAverageTime
. - Answers within
10-5
of the actual value will be accepted.
Implementation :
The first thing we should realize is that since checkins and checkouts are separated, we’ll need some type of data structure in which to store checkin information until we find the matching checkout information.
The next important realization is that since we only ultimately care about route times, we don’t really need to store checkout info at all. As long as we store the checkin info until we get to the checkout info, we can actually just then store the trip info separately by route and get rid of the checkin and checkout information. This will help keep the space needed to a minimum.
As we’ll want to look up checkin and route info by id and route name, we should use Map structures for both (checkins & routes). For the route information, we’ll only need to keep track of the number of trips and the total duration, so that we can calculate the average as needed. We can also use a concatenated name for the key in the route map in order to store the trip as a whole, rather than having to keep track of both ends separately.
For each user, store the checkin station and time.
For each trip (startStation + “_” + endStation), store the total time and counts.