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.
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)
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for uptofour people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before customers[i] arrive. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including beforeservingallcustomers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 12 * $1 - 4 * $92 = -$356.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
Solution: Simulation
Process if waiting customers > 0 or i < n.
Pruning, if runningCost > 4 * boardingCost (max revenue), there is no way to make profit.
Time complexity: sum(consumers) / 4 Space complexity: O(1)
Given an integer n and an integer array rounds. We have a circular track which consists of n sectors labeled from 1 to n. A marathon will be held on this track, the marathon consists of m rounds. The ith round starts at sector rounds[i - 1] and ends at sector rounds[i]. For example, round 1 starts at sector rounds[0] and ends at sector rounds[1]
Return an array of the most visited sectors sorted in ascending order.
Notice that you circulate the track in ascending order of sector numbers in the counter-clockwise direction (See the first example).
Example 1:
Input: n = 4, rounds = [1,3,1,2]
Output: [1,2]
Explanation: The marathon starts at sector 1. The order of the visited sectors is as follows:
1 --> 2 --> 3 (end of round 1) --> 4 --> 1 (end of round 2) --> 2 (end of round 3 and the marathon)
We can see that both sectors 1 and 2 are visited twice and they are the most visited sectors. Sectors 3 and 4 are visited only once.
Example 2:
Input: n = 2, rounds = [2,1,2,1,2,1,2,1,2]
Output: [2]
Example 3:
Input: n = 7, rounds = [1,3,5,7]
Output: [1,2,3,4,5,6,7]
Given an integer array arr of distinct integers and an integer k.
A game will be played between the first two elements of the array (i.e. arr[0] and arr[1]). In each round of the game, we compare arr[0] with arr[1], the larger integer wins and remains at position 0 and the smaller integer moves to the end of the array. The game ends when an integer wins k consecutive rounds.
Return the integer which will win the game.
It is guaranteed that there will be a winner of the game.
Example 1:
Input: arr = [2,1,3,5,4,6,7], k = 2
Output: 5
Explanation: Let's see the rounds of the game:
Round | arr | winner | win_count
1 | [2,1,3,5,4,6,7] | 2 | 1
2 | [2,3,5,4,6,7,1] | 3 | 1
3 | [3,5,4,6,7,1,2] | 5 | 1
4 | [5,4,6,7,1,2,3] | 5 | 2
So we can see that 4 rounds will be played and 5 is the winner because it wins 2 consecutive games.
Example 2:
Input: arr = [3,2,1], k = 10
Output: 3
Explanation: 3 will win the first 10 rounds consecutively.
Example 3:
Input: arr = [1,9,8,2,3,7,6,4,5], k = 7
Output: 9
Example 4:
Input: arr = [1,11,22,33,44,55,66,77,88,99], k = 1000000000
Output: 99
Since smaller numbers will be put to the end of the array, we must compare the rest of array before meeting those used numbers again. And the winner is monotonically increasing. So we can do it in one pass, just keep the largest number seen so far. If we reach to the end of the array, arr[0] will be max(arr) and it will always win no matter what k is.
