1396. Design Underground System
Description
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.
- A customer with a card ID equal to
void checkOut(int id, string stationName, int t)
- A customer with a card ID equal to
id
, checks out from the stationstationName
at timet
.
- A customer with a card ID equal to
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.
- Returns the average time it takes to travel from
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.
Solutions
Solution 1: Hash Table
We use two hash tables to store data:
ts
: Stores the passenger's id, check-in time, and check-in station. The key is the passenger's id, and the value is a tuple(t, stationName)
.d
: Stores the passenger's check-in station, check-out station, travel time, and number of trips. The key is a tuple(startStation, endStation)
, and the value is a tuple(totalTime, count)
.
When a passenger checks in, we store the passenger's id, check-in time, and check-in station in ts
, i.e., ts[id] = (t, stationName)
.
When a passenger checks out, we retrieve the passenger's check-in time and station (t0, station)
from ts
, then calculate the passenger's travel time $t - t_0$, and store the passenger's travel time and number of trips in d
.
When we want to calculate a passenger's average travel time, we retrieve the passenger's total travel time and number of trips (totalTime, count)
from d
, then calculate the average travel time as $totalTime / count$.
The time complexity is $O(1)$, and the space complexity is $O(n)$. Where $n$ is the number of passengers.
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