# 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]ExplanationUndergroundSystem 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 = 12undergroundSystem.checkOut(27, "Waterloo", 20);  // Customer 27 "Leyton" -> "Waterloo" in 20-10 = 10undergroundSystem.checkOut(32, "Cambridge", 22); // Customer 32 "Paradise" -> "Cambridge" in 22-8 = 14undergroundSystem.getAverageTime("Paradise", "Cambridge"); // return 14.00000. One trip "Paradise" -> "Cambridge", (14) / 1 = 14undergroundSystem.getAverageTime("Leyton", "Waterloo");    // return 11.00000. Two trips "Leyton" -> "Waterloo", (10 + 12) / 2 = 11undergroundSystem.checkIn(10, "Leyton", 24);undergroundSystem.getAverageTime("Leyton", "Waterloo");    // return 11.00000undergroundSystem.checkOut(10, "Waterloo", 38);  // Customer 10 "Leyton" -> "Waterloo" in 38-24 = 14undergroundSystem.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]ExplanationUndergroundSystem undergroundSystem = new UndergroundSystem();undergroundSystem.checkIn(10, "Leyton", 3);undergroundSystem.checkOut(10, "Paradise", 8); // Customer 10 "Leyton" -> "Paradise" in 8-3 = 5undergroundSystem.getAverageTime("Leyton", "Paradise"); // return 5.00000, (5) / 1 = 5undergroundSystem.checkIn(5, "Leyton", 10);undergroundSystem.checkOut(5, "Paradise", 16); // Customer 5 "Leyton" -> "Paradise" in 16-10 = 6undergroundSystem.getAverageTime("Leyton", "Paradise"); // return 5.50000, (5 + 6) / 2 = 5.5undergroundSystem.checkIn(2, "Leyton", 21);undergroundSystem.checkOut(2, "Paradise", 30); // Customer 2 "Leyton" -> "Paradise" in 30-21 = 9undergroundSystem.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.

# Implementation :

The first thing we should realize is that since checkins and checkouts are separated, we’ll need some type of data structure in which to store checkin information until we find the matching checkout information.

The next important realization is that since we only ultimately care about route times, we don’t really need to store checkout info at all. As long as we store the checkin info until we get to the checkout info, we can actually just then store the trip info separately by route and get rid of the checkin and checkout information. This will help keep the space needed to a minimum.

As we’ll want to look up checkin and route info by id and route name, we should use Map structures for both (checkins & routes). For the route information, we’ll only need to keep track of the number of trips and the total duration, so that we can calculate the average as needed. We can also use a concatenated name for the key in the route map in order to store the trip as a whole, rather than having to keep track of both ends separately.

For each user, store the checkin station and time.
For each trip (startStation + “_” + endStation), store the total time and counts.

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Full Stack Programmer, love to solve problem’s during free time.

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## Jaydeep

Full Stack Programmer, love to solve problem’s during free time.