1396. Design Underground System

Implement the class UndergroundSystem that supports three methods:

1.checkIn(int id, string stationName, int t)

• A customer with id card equal to id, gets in the station stationName at time t.

• A customer can only be checked into one place at a time.

• checkOut(int id, string stationName, int t)

• A customer with id card equal to id, gets out from the station stationName at time t.

• getAverageTime(string startStation, string endStation)

• Returns the average time to travel between the startStation and the endStation.

• The average time is computed from all the previous traveling from startStation to endStation that happened directly.
• Call to getAverageTime is always valid.

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.

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);
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

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);
undergroundSystem.getAverageTime("Leyton", "Paradise"); // return 5.00000
undergroundSystem.checkIn(5, "Leyton", 10);
undergroundSystem.checkOut(5, "Paradise", 16);
undergroundSystem.getAverageTime("Leyton", "Paradise"); // return 5.50000
undergroundSystem.checkIn(2, "Leyton", 21);
undergroundSystem.checkOut(2, "Paradise", 30);
undergroundSystem.getAverageTime("Leyton", "Paradise"); // return 6.66667

Constraints:

• There will be at most 20000 operations.
• 1 <= id, t <= 10^6
• All strings consist of uppercase, lowercase English letters and digits.
• 1 <= stationName.length <= 10
• Answers within 10^-5 of the actual value will be accepted as correct.

Analysis

The hard part for this problem is to get the average time quick. If we can have the route from startStation to endStation in $O(1)$ and also the total time of traveling in these two stations + total passengers in $O(1)$, then we can implement this function is $O(1)$.

We can create a map for storing the mentioned information: map<string, pair<int,int>> checkoutwhich the key is the route, and the value is (total traveling time, total passengers). If we want to get the result, we just need to return checkout[route].first / checkout[route].second;.

Now the problem becomes how to populate this map. Each id will be unique, so from this id we can know which station this user has begined from. Having this info we can populate the map when checkOut(int id, string stationName, int t), so that we need another map to store the relationship between the begin station associated with id and the timestamp.

• Time: $O(1)$
• Space: $O(number of routes + total passengers)$

Code

class UndergroundSystem {
public:
    UndergroundSystem() {

    }

    void checkIn(int id, string stationName, int t) {
        checkin[id] = {stationName, t};
    }

    void checkOut(int id, string stationName, int t) {
        string route = checkin[id].first + "-" + stationName;
        checkout[route].first += t - checkin[id].second;
        checkout[route].second ++;
    }

    double getAverageTime(string startStation, string endStation) {
        string route = startStation + "-" + endStation;
        return (double) checkout[route].first / checkout[route].second;
    }
private:
    unordered_map<int, pair<string,int>> checkin; // id -> (start station, checkin time)
    unordered_map<string, pair<int, int>> checkout; // route -> (total time, cnt)
};

/**
 * 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);
 */

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Last update: March 31, 2022