C++ ] leetCode 1396 - Design Underground System

 

 

리트코드 1396번 문제

 

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 station stationName at time t.
  • 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 station stationName at time t.
• double getAverageTime(string startStation, string endStation)
  • Returns the average time it takes to travel from startStation to endStation.
  • The average time is computed from all the previous traveling times from startStation to endStation that happened directly, meaning a check in at startStation followed by a check out from endStation.
  • The time it takes to travel from startStation to endStation may be different from the time it takes to travel from endStation to startStation.
  • There will be at least one customer that has traveled from startStation to endStation before getAverageTime 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 to checkIn, checkOut, and getAverageTime.
• Answers within 10-5 of the actual value will be accepted.

 

 

문제를 풀기 위에 map 변수를 두개 사용할건데 첫번째 map에는 <id, <startStation, checkInTime>을 저장하고 두번째 map에는 <<startStation, endStation>, <totalTravelTime, totalTravelCount>>를 저장한다.

 

손님이 체크아웃할 때마다 해당 경로에 대한 총 여행시간과 총 여행횟수를 업데이트하고 getAverageTime 함수가 호출됐을 때 평균값을 리턴하면 된다.

 

class UndergroundSystem {
public:
    map<int, pair<string, int>> customerInfo;
    map<pair<string,string>, pair<int, int>> travelTimes;
    UndergroundSystem() {
        
    }
    
    void checkIn(int id, string stationName, int t) {
        customerInfo[id] = make_pair(stationName, t);
    }
    
    void checkOut(int id, string stationName, int t) {
        auto it = customerInfo.find(id);
        if(it != customerInfo.end()) {
            travelTimes[make_pair(it->second.first, stationName)].first += t - it->second.second;
            travelTimes[make_pair(it->second.first, stationName)].second++;
            customerInfo.erase(id);
        }
    }
    
    double getAverageTime(string startStation, string endStation) {        
        auto it = travelTimes.find(make_pair(startStation, endStation));
        if(it != travelTimes.end()) return static_cast<double>(it->second.first) / it->second.second;
        else return 0;
    }
};