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1396. Design Underground System
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[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 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.
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댓글 2
mingki님의 댓글
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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.
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();
}
};
austin님의 댓글
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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;
};