KK's blog

每天积累多一些

0%

LeetCode 1396 Design Underground System

Posted on Disqus:

LeetCode



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 t<sub>1</sub> then checks out at time t<sub>2</sub>, then t<sub>1</sub> < t<sub>2</sub>. 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 <= 10<sup>6</sup> 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<sup>4</sup> calls in total to checkIn, checkOut, and getAverageTime.
* Answers within 10<sup>-5</sup> of the actual value will be accepted.

题目大意：

求两站之间的平均时间。checkin和checkout都会发生，一个人不能连续checkin两次。站都是按先到后。

解题思路：

用两个Map来记录customer id -> 站台和时间，另一个记录起始站pair -> 总距离多少trip pair

解题步骤：

N/A

注意事项：

  1. collections.defaultdict(lambda: [0, 0]) 用于value是pair

Python代码：

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
class Solution(TestCases):

    def __init__(self):
        self.start_to_end_time = {}
        self.start_end_to_total = collections.defaultdict(lambda: [0, 0])  # cant use defaultdict

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

    def checkOut(self, id: int, stationName: str, t: int) -> None:
        if id not in self.start_to_end_time:
            return
        (start_station, start_time) = self.start_to_end_time[id]
        total, n_trips = self.start_end_to_total[(start_station, stationName)]
        total += t - start_time
        n_trips += 1
        self.start_end_to_total[(start_station, stationName)] = (total, n_trips)

    def getAverageTime(self, startStation: str, endStation: str) -> float:
        total, n_trips = self.start_end_to_total[(startStation, endStation)]
        return total / n_trips

算法分析：

时间复杂度为O(1)，空间复杂度O(n)

Free mock interview