Design Underground System Python Solution

LeetCode 1396, Medium. Topics: Hash Table, String, Design. View on LeetCode. Generate this problem as a practice environment: tested reference solution, 13 parametrized pytest cases, and a playground notebook:

lcpy gen -n 1396   # by problem number
lcpy gen -s design_underground_system   # by problem name

Problem

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:

UndergroundSystem() Initializes the object of the underground system.
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.

Examples

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.checkIn(45, "Leyton", 3);
undergroundSystem.checkIn(32, "Paradise", 8);
undergroundSystem.checkIn(27, "Leyton", 10);
undergroundSystem.checkOut(45, "Waterloo", 15); // 15-3 = 12
undergroundSystem.checkOut(27, "Waterloo", 20); // 20-10 = 10
undergroundSystem.checkOut(32, "Cambridge", 22); // 22-8 = 14
undergroundSystem.getAverageTime("Paradise", "Cambridge"); // return 14.00000, (14) / 1 = 14
undergroundSystem.getAverageTime("Leyton", "Waterloo"); // return 11.00000, (10 + 12) / 2 = 11
undergroundSystem.checkIn(10, "Leyton", 24);
undergroundSystem.getAverageTime("Leyton", "Waterloo"); // return 11.00000
undergroundSystem.checkOut(10, "Waterloo", 38); // 38-24 = 14
undergroundSystem.getAverageTime("Leyton", "Waterloo"); // return 12.00000, (10 + 12 + 14) / 3 = 12

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.checkIn(10, "Leyton", 3);
undergroundSystem.checkOut(10, "Paradise", 8); // 8-3 = 5
undergroundSystem.getAverageTime("Leyton", "Paradise"); // return 5.00000, (5) / 1 = 5
undergroundSystem.checkIn(5, "Leyton", 10);
undergroundSystem.checkOut(5, "Paradise", 16); // 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); // 30-21 = 9
undergroundSystem.getAverageTime("Leyton", "Paradise"); // return 6.66667, (5 + 6 + 9) / 3 = 6.66667

Constraints

1 <= id, t <= 10^6
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 * 10^4 calls in total to checkIn, checkOut, and getAverageTime.
Answers within 10^-5 of the actual value will be accepted.

Solution

Reference implementation from solution.py on GitHub, full suite in test_solution.py:

class UndergroundSystem:
    # Time: check_in O(1), check_out O(1), get_average_time O(1)
    # Space: O(P + R) for P passengers in transit and R distinct routes
    def __init__(self) -> None:
        self.checked_in: dict[int, tuple[str, int]] = {}
        self.trips: dict[tuple[str, str], tuple[int, int]] = {}

    def check_in(self, id: int, station_name: str, t: int) -> None:
        self.checked_in[id] = (station_name, t)

    def check_out(self, id: int, station_name: str, t: int) -> None:
        start_station, start_t = self.checked_in.pop(id)
        route = (start_station, station_name)
        total, count = self.trips.get(route, (0, 0))
        self.trips[route] = (total + t - start_t, count + 1)

    def get_average_time(self, start_station: str, end_station: str) -> float:
        total, count = self.trips[(start_station, end_station)]
        return total / count

Complexity

Time	Space
check_in O(1), check_out O(1), get_average_time O(1)	O(P + R) for P passengers in transit and R distinct routes

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