Posts tagged as “simulation”

花花酱 1646. Get Maximum in Generated Array

By zxi on November 7, 2020

You are given an integer n. An array nums of length n + 1 is generated in the following way:

nums[0] = 0
nums[1] = 1
nums[2 * i] = nums[i] when 2 <= 2 * i <= n
nums[2 * i + 1] = nums[i] + nums[i + 1] when 2 <= 2 * i + 1 <= n

Returnthe maximum integer in the array nums.

Example 1:

Input: n = 7
Output: 3
Explanation: According to the given rules:
  nums[0] = 0
  nums[1] = 1
  nums[(1 * 2) = 2] = nums[1] = 1
  nums[(1 * 2) + 1 = 3] = nums[1] + nums[2] = 1 + 1 = 2
  nums[(2 * 2) = 4] = nums[2] = 1
  nums[(2 * 2) + 1 = 5] = nums[2] + nums[3] = 1 + 2 = 3
  nums[(3 * 2) = 6] = nums[3] = 2
  nums[(3 * 2) + 1 = 7] = nums[3] + nums[4] = 2 + 1 = 3
Hence, nums = [0,1,1,2,1,3,2,3], and the maximum is 3.

Example 2:

Input: n = 2
Output: 1
Explanation: According to the given rules, the maximum between nums[0], nums[1], and nums[2] is 1.

Example 3:

Input: n = 3
Output: 2
Explanation: According to the given rules, the maximum between nums[0], nums[1], nums[2], and nums[3] is 2.

Constraints:

0 <= n <= 100

Solution: Simulation

Generate the array by the given rules.

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

C++

class Solution {

public:

int getMaximumGenerated(int n) {

vector<int> nums(n + 1);

nums[0] = 0;

if (n > 0) nums[1] = 1;

for (int i = 1; i * 2 <= n; ++i) {

nums[2 * i] = nums[i];

if (i * 2 + 1 <= n) nums[2 * i + 1] = nums[i] + nums[i + 1];

}

return *max_element(begin(nums), end(nums));

}

};

花花酱 LeetCode 1606. Find Servers That Handled Most Number of Requests

By zxi on October 3, 2020

You have k servers numbered from 0 to k-1 that are being used to handle multiple requests simultaneously. Each server has infinite computational capacity but cannot handle more than one request at a time. The requests are assigned to servers according to a specific algorithm:

The i^th (0-indexed) request arrives.
If all servers are busy, the request is dropped (not handled at all).
If the (i % k)^th server is available, assign the request to that server.
Otherwise, assign the request to the next available server (wrapping around the list of servers and starting from 0 if necessary). For example, if the i^th server is busy, try to assign the request to the (i+1)^th server, then the (i+2)^th server, and so on.

You are given a strictly increasing array arrival of positive integers, where arrival[i] represents the arrival time of the i^th request, and another array load, where load[i] represents the load of the i^th request (the time it takes to complete). Your goal is to find the busiest server(s). A server is considered busiest if it handled the most number of requests successfully among all the servers.

Return a list containing the IDs (0-indexed) of the busiest server(s). You may return the IDs in any order.

Example 1:

Input: k = 3, arrival = [1,2,3,4,5], load = [5,2,3,3,3] 
Output: [1] 
Explanation:
All of the servers start out available.
The first 3 requests are handled by the first 3 servers in order.
Request 3 comes in. Server 0 is busy, so it's assigned to the next available server, which is 1.
Request 4 comes in. It cannot be handled since all servers are busy, so it is dropped.
Servers 0 and 2 handled one request each, while server 1 handled two requests. Hence server 1 is the busiest server.

Example 2:

Input: k = 3, arrival = [1,2,3,4], load = [1,2,1,2]
Output: [0]
Explanation:
The first 3 requests are handled by first 3 servers.
Request 3 comes in. It is handled by server 0 since the server is available.
Server 0 handled two requests, while servers 1 and 2 handled one request each. Hence server 0 is the busiest server.

Example 3:

Input: k = 3, arrival = [1,2,3], load = [10,12,11]
Output: [0,1,2]
Explanation: Each server handles a single request, so they are all considered the busiest.

Example 4:

Input: k = 3, arrival = [1,2,3,4,8,9,10], load = [5,2,10,3,1,2,2]
Output: [1]

Example 5:

Input: k = 1, arrival = [1], load = [1]
Output: [0]

Constraints:

1 <= k <= 10⁵
1 <= arrival.length, load.length <= 10⁵
arrival.length == load.length
1 <= arrival[i], load[i] <= 10⁹
arrival is strictly increasing.

Solution: Heap + TreeSet

Use a min heap to store the release time -> server.
Use a treeset to track the current available servers.
For reach request, check whether servers can be released at that time.

Time complexity: O(nlogk)
Space complexity: O(k)

C++

// Author: Huahua

class Solution {

public:

vector<int> busiestServers(int k, vector<int>& arrival, vector<int>& load) {

priority_queue<pair<int, int>, vector<pair<int,int>>, greater<>> q; // {release_time, server}

set<int> servers;

vector<int> requests(k);

for (int i = 0; i < k; ++i)

servers.insert(i);

for (int i = 0; i < arrival.size(); ++i) {

const int t = arrival[i];

const int l = load[i];

// Release servers. O(logk) per pop()

while (!q.empty() && q.top().first <= t) {

servers.insert(q.top().second);

q.pop();

}

// Drop the request.

if (servers.empty()) continue;

// Find first avaiable one O(logk)

auto it = servers.lower_bound(i % k);

if (it == servers.end()) it = begin(servers);

const int idx = *it;

++requests[idx];

servers.erase(it);

q.emplace(t + l, idx);

}

const int max_req = *max_element(begin(requests), end(requests));

vector<int> ans;

for (int i = 0; i < k; ++i)

if (requests[i] == max_req) ans.push_back(i);

return ans;

}

};

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 1599. Maximum Profit of Operating a Centennial Wheel

By zxi on September 27, 2020

You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four 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 i^th 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 before serving all customers. 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)

C++

class Solution {

public:

int minOperationsMaxProfit(vector<int>& customers, int boardingCost, int runningCost) {

const int n = customers.size();

constexpr int kMaxC = 4;

if (runningCost > kMaxC * boardingCost) return -1;

int c = 0;

int ans = -1;

int p = 0;

int max_p = 0;

for (int r = 0; r < n || c > 0; ++r) {

c += (r < n ? customers[r] : 0);

p += min(c, kMaxC) * boardingCost - runningCost;

c -= min(c, kMaxC);

if (p > max_p) {

max_p = p;

ans = r + 1; // 1-based

}

return ans;

}

};

花花酱 LeetCode 1598. Crawler Log Folder

By zxi on September 27, 2020

The Leetcode file system keeps a log each time some user performs a change folder operation.

The operations are described below:

"../" : Move to the parent folder of the current folder. (If you are already in the main folder, remain in the same folder).
"./" : Remain in the same folder.
"x/" : Move to the child folder named x (This folder is guaranteed to always exist).

You are given a list of strings logs where logs[i] is the operation performed by the user at the i^th step.

The file system starts in the main folder, then the operations in logs are performed.

Return the minimum number of operations needed to go back to the main folder after the change folder operations.

Example 1:

Input: logs = ["d1/","d2/","../","d21/","./"]
Output: 2
Explanation: Use this change folder operation "../" 2 times and go back to the main folder.

Example 2:

Input: logs = ["d1/","d2/","./","d3/","../","d31/"]
Output: 3

Example 3:

Input: logs = ["d1/","../","../","../"]
Output: 0

Constraints:

1 <= logs.length <= 10³
2 <= logs[i].length <= 10
logs[i] contains lowercase English letters, digits, '.', and '/'.
logs[i] follows the format described in the statement.
Folder names consist of lowercase English letters and digits.

Solution: Simulation

We only need to track the depth of current folder, and name and path can be ignored.

Time complexity: O(n)
Space complexity: O(1)

C++

// Author: Huahua

class Solution {

public:

int minOperations(vector<string>& logs) {

int ans = 0;

for (const string& log : logs) {

if (log == "../")

ans = max(ans - 1, 0);

else if (log != "./")

++ans;

}

return ans;

}

};