LeetCode 솔루션

1396. Design Underground System

mingki 쪽지보내기 메일보내기 자기소개 아이디로 검색 전체게시물 작성
작성일 2022.04.28 18:55

17,084 조회
2 댓글
0 추천
0 비추천
목록

글검색

게시글 쓰기

[LeetCode 시즌 3] 2022년 4월 23일 문제입니다.

https://leetcode.com/problems/design-underground-system/

1396. Design Underground System

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.

#LeetCode, #Bloomberg, #Hash Table, #String, #Design

링크

https://leetcode.com/problems/design-underground-system/ 4282 회 연결

익명
작성일 2022.04.28 19:00

C++
Runtime: 172 ms, faster than 76.43% of C++ online submissions for Design Underground System.
Memory Usage: 59.9 MB, less than 16.30% of C++ online submissions for Design Underground System.

class UndergroundSystem {
private:
    struct Info {
        string stationName;
        int time;
    };
    
    unordered_map<int, Info> checked;
    unordered_map<string, vector<int>> distance;
    
public:
    UndergroundSystem() {
        
    }
    
    void checkIn(int id, string stationName, int t) {
        checked[id].stationName = stationName;
        checked[id].time = t;
    }
    
    void checkOut(int id, string stationName, int t) {
        string fromTo = checked[id].stationName + "." + stationName;
        distance[fromTo].push_back(t - checked[id].time);
    }
    
    double getAverageTime(string startStation, string endStation) {
        string fromTo = startStation + "." + endStation;
        return 1.0 * accumulate(distance[fromTo].begin(), distance[fromTo].end(), 0) / distance[fromTo].size();
    }
};

C++
Runtime: 172 ms, faster than 76.43% of C++ online submissions for Design Underground System.
Memory Usage: 59.9 MB, less than 16.30% of C++ online submissions for Design Underground System.
[code=C++]class UndergroundSystem {
private:
    struct Info {
        string stationName;
        int time;
    };
    
    unordered_map<int, Info> checked;
    unordered_map<string, vector<int>> distance;
    
public:
    UndergroundSystem() {
        
    }
    
    void checkIn(int id, string stationName, int t) {
        checked[id].stationName = stationName;
        checked[id].time = t;
    }
    
    void checkOut(int id, string stationName, int t) {
        string fromTo = checked[id].stationName + "." + stationName;
        distance[fromTo].push_back(t - checked[id].time);
    }
    
    double getAverageTime(string startStation, string endStation) {
        string fromTo = startStation + "." + endStation;
        return 1.0 * accumulate(distance[fromTo].begin(), distance[fromTo].end(), 0) / distance[fromTo].size();
    }
};[/code]

익명
작성일 2022.04.29 22:08

C++ All const time complexity solution

class UndergroundSystem {
public:
    UndergroundSystem() {
        
    }
    
    void checkIn(int id, string stationName, int t) {
        p[id] = tuple(stationName, t);
    }
    
    void checkOut(int id, string stationName, int t) {
        auto [startStation, checkInTime] = p[id];
        auto &[sum, count] = m[startStation][stationName];
        ++count;
        sum += t-checkInTime;
    }
    
    double getAverageTime(string startStation, string endStation) {
        auto &[sum, count] = m[startStation][endStation];
        return sum / count;
    }
    unordered_map<string, unordered_map<string, pair<double, int>>> m;
    unordered_map<int, tuple<string, int>> p;
};