1396. Design Underground System

https://leetcode.com/problems/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.

2. checkOut(int id, string stationName, int t)

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

3. 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,14.0,11.0,null,11.0,null,12.0]

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 14.0. There was only one travel from "Paradise" (at time 8) to "Cambridge" (at time 22)
undergroundSystem.getAverageTime("Leyton", "Waterloo");          // return 11.0. 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 = 11.0
undergroundSystem.checkIn(10, "Leyton", 24);
undergroundSystem.getAverageTime("Leyton", "Waterloo");          // return 11.0
undergroundSystem.checkOut(10, "Waterloo", 38);
undergroundSystem.getAverageTime("Leyton", "Waterloo");          // return 12.0

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.

设计简易进出站统计系统，分别实现：checkIn(t, s, u)表示用户u在t时刻进s站，checkOut(t, s, u)表示出站，getAvgTime(s1, s2)统计所有直接从s1到s2站的旅客平均花费时间。需要统计s1到s2的旅客平均时间，因此用dict s[s1][s2]记录从s1到s2的总时间和总人数。再用dict u记录用户何时进出何站。

class UndergroundSystem:

    def __init__(self):
        self.s, self.u = {}, {}
        
    def checkIn(self, id: int, sn: str, t: int) -> None:
        self.u[id] = [sn, t] 
        
    def checkOut(self, id: int, sn: str, t: int) -> None:
        s, u = self.s, self.u
        st, t0 = u[id][0], u[id][1]
        if st not in s:
            s[st] = {}
        if sn not in s[st]:
            # (total time, total users)
            s[st][sn] = [0, 0]
        s[st][sn][0] += t - t0
        s[st][sn][1] += 1
        u[id] = None

    def getAverageTime(self, start: str, end: str) -> float:
        return float(self.s[start][end][0]) / float(self.s[start][end][1])


# Your UndergroundSystem object will be instantiated and called as such:
# obj = UndergroundSystem()
# obj.checkIn(id,stationName,t)
# obj.checkOut(id,stationName,t)
# param_3 = obj.getAverageTime(startStation,endStation)