1396. Design Underground System LeetCode Solution

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Problem Statement
Complexity Analysis
Design Underground System solution in C++
Design Underground System solution in Java
Design Underground System solution in Python
Additional Resources

1396. Design Underground System LeetCode Solution image

Problem Statement of 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.

Complexity Analysis

Time Complexity: O(1)
Space Complexity: O(|\texttt{passengers}| + |\texttt{stations}|^2)

1396. Design Underground System LeetCode Solution in C++

struct CheckIn {
  string stationName;
  int time;
};

struct CheckOut {
  int numTrips;
  int totalTime;
};

class UndergroundSystem {
 public:
  void checkIn(int id, string stationName, int t) {
    checkIns[id] = {stationName, t};
  }

  void checkOut(int id, string stationName, int t) {
    const auto [startStation, startTime] = checkIns[id];
    checkIns.erase(id);
    const string& route = startStation + "->" + stationName;
    ++checkOuts[route].numTrips;
    checkOuts[route].totalTime += t - startTime;
  }

  double getAverageTime(string startStation, string endStation) {
    const auto& [numTrips, totalTime] =
        checkOuts[startStation + "->" + endStation];
    return totalTime / (double)numTrips;
  }

 private:
  unordered_map<int, CheckIn> checkIns;       // {id: (stationName, time)}
  unordered_map<string, CheckOut> checkOuts;  // {route: (numTrips, totalTime)}
};
/* code provided by PROGIEZ */

1396. Design Underground System LeetCode Solution in Java

class CheckIn {
  public String stationName;
  public int time;
  public CheckIn(String stationName, int time) {
    this.stationName = stationName;
    this.time = time;
  }
}

class CheckOut {
  public int numTrips;
  public int totalTime;
  public CheckOut(int numTrips, int totalTime) {
    this.numTrips = numTrips;
    this.totalTime = totalTime;
  }
}

class UndergroundSystem {
  public void checkIn(int id, String stationName, int t) {
    checkIns.put(id, new CheckIn(stationName, t));
  }

  public void checkOut(int id, String stationName, int t) {
    final CheckIn checkIn = checkIns.get(id);
    checkIns.remove(id);
    final String route = checkIn.stationName + "->" + stationName;
    checkOuts.putIfAbsent(route, new CheckOut(0, 0));
    ++checkOuts.get(route).numTrips;
    checkOuts.get(route).totalTime += t - checkIn.time;
  }

  public double getAverageTime(String startStation, String endStation) {
    final CheckOut checkOut = checkOuts.get(startStation + "->" + endStation);
    return checkOut.totalTime / (double) checkOut.numTrips;
  }

  private Map<Integer, CheckIn> checkIns = new HashMap<>();  // {id: (stationName, time)}
  private Map<String, CheckOut> checkOuts = new HashMap<>(); // {route: (numTrips, totalTime)}
}
// code provided by PROGIEZ

1396. Design Underground System LeetCode Solution in Python

class UndergroundSystem:
  def __init__(self):
    # {id: (stationName, time)}
    self.checkIns = {}
    # {route: (numTrips, totalTime)}
    self.checkOuts = collections.defaultdict(lambda: [0, 0])

  def checkIn(self, id: int, stationName: str, t: int) -> None:
    self.checkIns[id] = (stationName, t)

  def checkOut(self, id: int, stationName: str, t: int) -> None:
    startStation, startTime = self.checkIns.pop(id)
    route = (startStation, stationName)
    self.checkOuts[route][0] += 1
    self.checkOuts[route][1] += t - startTime

  def getAverageTime(self, startStation: str, endStation: str) -> float:
    numTrips, totalTime = self.checkOuts[(startStation, endStation)]
    return totalTime / numTrips
 # code by PROGIEZ