g1301_1400.s1396_design_underground_system.UndergroundSystem Maven / Gradle / Ivy

Go to download
Show more of this group Show more artifacts with this name
Show all versions of leetcode-in-java21 Show documentation
Java-based LeetCode algorithm problem solutions, regularly updated
There is a newer version: 1.38
package g1301_1400.s1396_design_underground_system;

// #Medium #String #Hash_Table #Design #2022_03_17_Time_89_ms_(89.64%)_Space_55.2_MB_(88.83%)

import java.util.HashMap;
import java.util.LinkedList;
import java.util.Map;

/**
 * 1396 - Design Underground System\.
 *
 * Medium
 *
 * 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.
**/
public class UndergroundSystem {
    private static class StationAndTime {
        String stationName;
        int t;

        public StationAndTime(String stationName, int t) {
            this.stationName = stationName;
            this.t = t;
        }

        public String getStation() {
            return this.stationName;
        }

        public int getTime() {
            return this.t;
        }
    }

    private Map averageTimeMap;
    private Map> travelerMap;

    public UndergroundSystem() {
        averageTimeMap = new HashMap<>();
        travelerMap = new HashMap<>();
    }

    public void checkIn(int id, String stationName, int t) {
        travelerMap.putIfAbsent(id, new LinkedList<>());
        travelerMap.get(id).add(new StationAndTime(stationName, t));
    }

    public void checkOut(int id, String stationName, int t) {
        LinkedList list = travelerMap.get(id);
        StationAndTime stationAndTime = list.getLast();
        String startToEndStation = stationAndTime.getStation() + "->" + stationName;
        int duration = t - stationAndTime.getTime();
        if (averageTimeMap.containsKey(startToEndStation)) {
            double[] pair = averageTimeMap.get(startToEndStation);
            double newAverage = (pair[0] * pair[1] + duration) / (pair[1] + 1);
            averageTimeMap.put(startToEndStation, new double[] {newAverage, pair[1] + 1});
        } else {
            averageTimeMap.put(startToEndStation, new double[] {duration, 1});
        }
    }

    public double getAverageTime(String startStation, String endStation) {
        return averageTimeMap.get(startStation + "->" + endStation)[0];
    }
}

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