How do you solve Design Underground System on LeetCode?

Design Underground System (LeetCode #1396) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(1) for checkIn and checkOut, O(1) for getAverageTime time complexity and O(n) for storing check-ins and travel times space complexity. Key insight: Using hash maps allows for efficient storage and retrieval of data, which is crucial for performance in this problem.

What is the time complexity of Design Underground System?

The optimal solution for Design Underground System runs in O(1) for checkIn and checkOut, O(1) for getAverageTime time and uses O(n) for storing check-ins and travel times space. The time complexity is O(1) for both checkIn and checkOut operations because we are using hash maps for constant time access. The getAverageTime operation is also O(1) as it retrieves pre-computed totals and counts.

What is the brute force solution for Design Underground System?

The brute force approach for Design Underground System is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves storing all check-in and check-out times and calculating the average time for each query by iterating through all previous records. This is simple but inefficient as it requires checking all records every time we want to calculate an average. The optimal Optimal Solution approach improves this to O(1) for checkIn and checkOut, O(1) for getAverageTime.

What algorithm or data structure is used to solve Design Underground System?

Design Underground System uses the following concepts: Hash Table, String, Design. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Design Underground System?

Clearly explain your thought process and why you choose a particular data structure. Consider edge cases and how your solution handles them. Optimize your solution step-by-step, explaining the trade-offs of each approach.

What are common mistakes when solving Design Underground System?

Not properly managing the check-in and check-out states, leading to incorrect travel time calculations. Failing to handle the case where a route has not been traveled yet, which can lead to division by zero.

#1396

Design Underground System

Medium

Hash TableStringDesignHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(1) for checkIn and checkOut, O(1) for getAverageTime
Space	O(1)	O(n) for storing check-ins and travel times

💡

Intuition

Time O(1) for checkIn and checkOut, O(1) for getAverageTimeSpace O(n) for storing check-ins and travel times

The optimal solution uses two hash maps: one to store the check-in data and another to store cumulative travel times and counts for each route. This allows us to compute averages in constant time after the initial check-in and check-out operations.

⚙️

Algorithm

5 steps

1Step 1: Create a hash map to store check-in data with customer ID as the key.
2Step 2: Create another hash map to store cumulative travel times and counts for each route (startStation-endStation).
3Step 3: For checkIn, store the station and time in the first hash map.
4Step 4: For checkOut, calculate the travel time, update the cumulative travel time and count in the second hash map.
5Step 5: For getAverageTime, retrieve the total travel time and count for the route and compute the average.

solution.py22 lines

1class UndergroundSystem:
2    def __init__(self):
3        self.checkins = {}
4        self.travel_times = {}
5        self.counts = {}
6
7    def checkIn(self, id: int, stationName: str, t: int) -> None:
8        self.checkins[id] = (stationName, t)
9
10    def checkOut(self, id: int, stationName: str, t: int) -> None:
11        station, checkin_time = self.checkins.pop(id)
12        travel_time = t - checkin_time
13        route = (station, stationName)
14        if route not in self.travel_times:
15            self.travel_times[route] = 0
16            self.counts[route] = 0
17        self.travel_times[route] += travel_time
18        self.counts[route] += 1
19
20    def getAverageTime(self, startStation: str, endStation: str) -> float:
21        route = (startStation, endStation)
22        return self.travel_times[route] / self.counts[route]

ℹ

Complexity note: The time complexity is O(1) for both checkIn and checkOut operations because we are using hash maps for constant time access. The getAverageTime operation is also O(1) as it retrieves pre-computed totals and counts.

1Using hash maps allows for efficient storage and retrieval of data, which is crucial for performance in this problem.
2Maintaining separate counts and total travel times for each route enables quick average calculations.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.