underground-system py3
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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.
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Implement the `UndergroundSystem` class:
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* `void checkIn(int id, string stationName, int t)`
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* A customer with a card ID equal to `id`, checks in at the station `stationName` at time `t`.
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* A customer can only be checked into one place at a time.
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* `void checkOut(int id, string stationName, int t)`
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* A customer with a card ID equal to `id`, checks out from the station `stationName` at time `t`.
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* `double getAverageTime(string startStation, string endStation)`
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* Returns the average time it takes to travel from `startStation` to `endStation`.
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* 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`.
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* The time it takes to travel from `startStation` to `endStation` **may be different** from the time it takes to travel from `endStation` to `startStation`.
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* There will be at least one customer that has traveled from `startStation` to `endStation` before `getAverageTime` is called.
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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.
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**Example 1:**
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Input
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["UndergroundSystem","checkIn","checkIn","checkIn","checkOut","checkOut","checkOut","getAverageTime","getAverageTime","checkIn","getAverageTime","checkOut","getAverageTime"]
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[[],[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"]]
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Output
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[null,null,null,null,null,null,null,14.00000,11.00000,null,11.00000,null,12.00000]
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Explanation
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UndergroundSystem undergroundSystem = new UndergroundSystem();
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undergroundSystem.checkIn(45, "Leyton", 3);
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undergroundSystem.checkIn(32, "Paradise", 8);
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undergroundSystem.checkIn(27, "Leyton", 10);
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undergroundSystem.checkOut(45, "Waterloo", 15); // Customer 45 "Leyton" -> "Waterloo" in 15-3 = 12
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undergroundSystem.checkOut(27, "Waterloo", 20); // Customer 27 "Leyton" -> "Waterloo" in 20-10 = 10
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undergroundSystem.checkOut(32, "Cambridge", 22); // Customer 32 "Paradise" -> "Cambridge" in 22-8 = 14
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undergroundSystem.getAverageTime("Paradise", "Cambridge"); // return 14.00000. One trip "Paradise" -> "Cambridge", (14) / 1 = 14
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undergroundSystem.getAverageTime("Leyton", "Waterloo"); // return 11.00000. Two trips "Leyton" -> "Waterloo", (10 + 12) / 2 = 11
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undergroundSystem.checkIn(10, "Leyton", 24);
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undergroundSystem.getAverageTime("Leyton", "Waterloo"); // return 11.00000
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undergroundSystem.checkOut(10, "Waterloo", 38); // Customer 10 "Leyton" -> "Waterloo" in 38-24 = 14
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undergroundSystem.getAverageTime("Leyton", "Waterloo"); // return 12.00000. Three trips "Leyton" -> "Waterloo", (10 + 12 + 14) / 3 = 12
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**Example 2:**
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Input
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["UndergroundSystem","checkIn","checkOut","getAverageTime","checkIn","checkOut","getAverageTime","checkIn","checkOut","getAverageTime"]
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[[],[10,"Leyton",3],[10,"Paradise",8],["Leyton","Paradise"],[5,"Leyton",10],[5,"Paradise",16],["Leyton","Paradise"],[2,"Leyton",21],[2,"Paradise",30],["Leyton","Paradise"]]
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Output
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[null,null,null,5.00000,null,null,5.50000,null,null,6.66667]
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Explanation
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UndergroundSystem undergroundSystem = new UndergroundSystem();
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undergroundSystem.checkIn(10, "Leyton", 3);
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undergroundSystem.checkOut(10, "Paradise", 8); // Customer 10 "Leyton" -> "Paradise" in 8-3 = 5
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undergroundSystem.getAverageTime("Leyton", "Paradise"); // return 5.00000, (5) / 1 = 5
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undergroundSystem.checkIn(5, "Leyton", 10);
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undergroundSystem.checkOut(5, "Paradise", 16); // Customer 5 "Leyton" -> "Paradise" in 16-10 = 6
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undergroundSystem.getAverageTime("Leyton", "Paradise"); // return 5.50000, (5 + 6) / 2 = 5.5
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undergroundSystem.checkIn(2, "Leyton", 21);
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undergroundSystem.checkOut(2, "Paradise", 30); // Customer 2 "Leyton" -> "Paradise" in 30-21 = 9
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undergroundSystem.getAverageTime("Leyton", "Paradise"); // return 6.66667, (5 + 6 + 9) / 3 = 6.66667
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**Constraints:**
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* `1 <= id, t <= 106`
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* `1 <= stationName.length, startStation.length, endStation.length <= 10`
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* All strings consist of uppercase and lowercase English letters and digits.
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* There will be at most `2 * 104` calls **in total** to `checkIn`, `checkOut`, and `getAverageTime`.
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* Answers within `10-5` of the actual value will be accepted.
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https://leetcode.com/problems/design-underground-system
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# Time: O(N)
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# Space: O(P) P is num of passengers who checkin at same time worst case
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# + O(S^2) S is num of stations, pair up every station so S^2 permutation
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class UndergroundSystem:
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def __init__(self):
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self.checkins = {}
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self.totals = {}
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def checkIn(self, id: int, stationName: str, t: int) -> None:
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self.checkins[id] = (stationName, t)
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def checkOut(self, id: int, endStation: str, end_t: int) -> None:
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startStation, start_t = self.checkins[id]
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key = (startStation, endStation)
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duration = end_t - start_t
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curr_total = self.totals.get(key, (0, 0))
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# Increment total and count of journeys
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self.totals[key] = (curr_total[0] + duration, curr_total[1] + 1)
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# Journey is over for this passenger, we can remove it since we
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# tallied the data
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del self.checkins[id]
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def getAverageTime(self, startStation: str, endStation: str) -> float:
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total_duration, num_trips = self.totals[(startStation, endStation)]
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return total_duration / num_trips
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# Your UndergroundSystem object will be instantiated and called as such:
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# obj = UndergroundSystem()
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# obj.checkIn(id,stationName,t)
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# obj.checkOut(id,stationName,t)
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# param_3 = obj.getAverageTime(startStation,endStation)
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