花花酱 LeetCode 1603. Design Parking System

By zxi on October 3, 2020

Design a parking system for a parking lot. The parking lot has three kinds of parking spaces: big, medium, and small, with a fixed number of slots for each size.

Implement the ParkingSystem class:

ParkingSystem(int big, int medium, int small) Initializes object of the ParkingSystem class. The number of slots for each parking space are given as part of the constructor.
bool addCar(int carType) Checks whether there is a parking space of carType for the car that wants to get into the parking lot. carType can be of three kinds: big, medium, or small, which are represented by 1, 2, and 3 respectively. A car can only park in a parking space of its carType. If there is no space available, return false, else park the car in that size space and return true.

Example 1:

Input
["ParkingSystem", "addCar", "addCar", "addCar", "addCar"]
[[1, 1, 0], [1], [2], [3], [1]]
Output
[null, true, true, false, false]

Explanation ParkingSystem parkingSystem = new ParkingSystem(1, 1, 0); parkingSystem.addCar(1); // return true because there is 1 available slot for a big car parkingSystem.addCar(2); // return true because there is 1 available slot for a medium car parkingSystem.addCar(3); // return false because there is no available slot for a small car parkingSystem.addCar(1); // return false because there is no available slot for a big car. It is already occupied.

Constraints:

0 <= big, medium, small <= 1000
carType is 1, 2, or 3
At most 1000 calls will be made to addCar

Solution: Simulation

Time complexity: O(1) per addCar call
Space complexity: O(1)

C++

// Author: Huahua

class ParkingSystem {

public:

ParkingSystem(int big, int medium, int small):

slots_{{big, medium, small}} {}

bool addCar(int carType) {

return slots_[carType - 1]-- > 0;

}

private:

vector<int> slots_;

};

Python3

# Author: Huahua

class ParkingSystem:

def __init__(self, big: int, medium: int, small: int):

self.slots = {1: big, 2: medium, 3: small}

def addCar(self, carType: int) -> bool:

if self.slots[carType] == 0: return False

self.slots[carType] -= 1

return True

花花酱 LeetCode 1396. Design Underground System

By zxi on March 28, 2020

Implement the class UndergroundSystem that supports three methods:

1. checkIn(int id, string stationName, int t)

A customer with id card equal to id, gets in the station stationName at time t.
A customer can only be checked into one place at a time.

2. checkOut(int id, string stationName, int t)

A customer with id card equal to id, gets out from the station stationName at time t.

3. getAverageTime(string startStation, string endStation)

Returns the average time to travel between the startStation and the endStation.
The average time is computed from all the previous traveling from startStation to endStation that happened directly.
Call to getAverageTime is always valid.

You can assume all calls to checkIn and checkOut methods are consistent. That is, if a customer gets in at time t₁ at some station, then it gets out at time t₂ with t₂ > t₁. 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,14.0,11.0,null,11.0,null,12.0]

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);
undergroundSystem.checkOut(27, "Waterloo", 20);
undergroundSystem.checkOut(32, "Cambridge", 22);
undergroundSystem.getAverageTime("Paradise", "Cambridge");       // return 14.0. There was only one travel from "Paradise" (at time 8) to "Cambridge" (at time 22)
undergroundSystem.getAverageTime("Leyton", "Waterloo");          // return 11.0. There were two travels from "Leyton" to "Waterloo", a customer with id=45 from time=3 to time=15 and a customer with id=27 from time=10 to time=20. So the average time is ( (15-3) + (20-10) ) / 2 = 11.0
undergroundSystem.checkIn(10, "Leyton", 24);
undergroundSystem.getAverageTime("Leyton", "Waterloo");          // return 11.0
undergroundSystem.checkOut(10, "Waterloo", 38);
undergroundSystem.getAverageTime("Leyton", "Waterloo");          // return 12.0

Constraints:

There will be at most 20000 operations.
1 <= id, t <= 10^6
All strings consist of uppercase, lowercase English letters and digits.
1 <= stationName.length <= 10
Answers within 10^-5 of the actual value will be accepted as correct.

Solution: Hashtable

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

Time complexity: O(n)
Space complexity: O(n)

C++

// Author: Huahua

class UndergroundSystem {

public:

UndergroundSystem() {}

void checkIn(int id, string stationName, int t) {

m_[id] = {stationName, t};

}

void checkOut(int id, string stationName, int t) {

const auto& [s0, t0] = m_[id];

string key = s0 + "_" + stationName;

times_[key].first += (t - t0);

++times_[key].second;

}

double getAverageTime(string startStation, string endStation) {

const auto& [sum, count] = times_[startStation + "_" + endStation];

return static_cast<double>(sum) / count;

}

private:

unordered_map<int, pair<string, int>> m_; // id -> {station, t}

unordered_map<string, pair<int, int>> times_; // trip -> {sum, count}

};

花花酱 LeetCode 592. Fraction Addition and Subtraction

By zxi on August 6, 2018

Problem

Given a string representing an expression of fraction addition and subtraction, you need to return the calculation result in string format. The final result should be irreducible fraction. If your final result is an integer, say 2, you need to change it to the format of fraction that has denominator 1. So in this case, 2 should be converted to 2/1.

Example 1:

Input:"-1/2+1/2"
Output: "0/1"

Example 2:

Input:"-1/2+1/2+1/3"
Output: "1/3"

Example 3:

Input:"1/3-1/2"
Output: "-1/6"

Example 4:

Input:"5/3+1/3"
Output: "2/1"

Note:

The input string only contains '0' to '9', '/', '+' and '-'. So does the output.
Each fraction (input and output) has format ±numerator/denominator. If the first input fraction or the output is positive, then '+' will be omitted.
The input only contains valid irreducible fractions, where the numerator and denominator of each fraction will always be in the range [1,10]. If the denominator is 1, it means this fraction is actually an integer in a fraction format defined above.
The number of given fractions will be in the range [1,10].
The numerator and denominator of the final result are guaranteed to be valid and in the range of 32-bit int.

Solution: Math

a/b+c/d = (a*d + b * c) / (b * d)

Time complexity: O(n)

Space complexity: O(1)

C++

// Author: Huahua

class Solution {

public:

string fractionAddition(string expression) {

int a = 0;

int b = 1;

int c;

int d;

char slash;

istringstream in(expression);

while (in >> c >> slash >> d) {

a = a * d + b * c;

b *= d;

int gcd = abs(__gcd(a, b));

a /= gcd;

b /= gcd;

}

return to_string(a) + "/" + to_string(b);

}

};

C++/class

// Author: Huahua

// Running time: 0 ms

class Fraction {

public:

Fraction(int a, int b): a_(a), b_(b) {}

Fraction& operator+=(const Fraction& f) {

a_ = a_ * f.b_ + b_ * f.a_;

b_ = b_ * f.b_;

int d = abs(__gcd(a_, b_));

a_ /= d;

b_ /= d;

return *this;

}

string toString() {

return to_string(a_) + "/" + to_string(b_);

}

private:

int a_; // hold the sign, e.g. can be +-.

int b_; // always > 0.

};

class Solution {

public:

string fractionAddition(string expression) {

Fraction f(0, 1);

int a;

int b;

char op;

istringstream in(expression);

while (in >> a >> op >> b)

f += Fraction(a, b);

return f.toString();

}

};

Posts tagged as “class”

花花酱 LeetCode 1603. Design Parking System

Solution: Simulation

C++

Python3

花花酱 LeetCode 1396. Design Underground System

Solution: Hashtable

C++

花花酱 LeetCode 592. Fraction Addition and Subtraction

Problem

Solution: Math

C++

C++/class