Leetcode 1882: Process Tasks Using Servers
You are given two integer arrays,
servers and tasks. The servers array represents the weights of the servers, and the tasks array represents the processing times of tasks. Tasks are assigned to servers based on the smallest server weight and index. If no server is available, tasks wait for a server to become free.Problem
Approach
Steps
Complexity
Input: You are given two arrays: `servers` and `tasks`. The `servers` array is a list of integers representing the weight of each server. The `tasks` array contains integers representing the processing times for tasks. At each second, a new task enters the queue and is assigned to the server based on availability and weight priorities.
Example: servers = [5, 1, 4, 3, 2], tasks = [2, 1, 2, 4, 5, 2, 1]
Constraints:
• 1 <= n, m <= 2 * 10^5
• 1 <= servers[i], tasks[j] <= 2 * 10^5
Output: You need to return an array of indices representing which server each task is assigned to. The output should follow the task queue order.
Example: [1, 4, 1, 4, 1, 3, 2]
Constraints:
• The output array will have the same length as the `tasks` array.
Goal: To simulate task assignments to servers based on availability, weight, and index priority.
Steps:
• Initialize a priority queue for free servers based on their weights and indices.
• For each task, check if there are any free servers available. If so, assign the task to the one with the smallest weight and index.
• If no servers are available, check when the next server becomes free and assign tasks accordingly.
• Keep track of task assignments and update the server statuses (free or busy).
Goal: The constraints specify the maximum number of servers and tasks that can be handled. Be mindful of large inputs and ensure the solution can handle them efficiently.
Steps:
• servers.length == n
• tasks.length == m
• 1 <= n, m <= 2 * 10^5
• 1 <= servers[i], tasks[j] <= 2 * 10^5
Assumptions:
• Each server is initially free and available to process tasks.
• Tasks are processed in the order they are added to the queue.
• Servers will be used based on availability, weight, and index priority.
• Input: Example 1
• Explanation: In the first example, we see that tasks are assigned based on server weights. Server 1 (weight 4) is assigned task 0, and the process continues in this manner for each task in the queue. Servers that become free are used for subsequent tasks.
• Input: Example 2
• Explanation: In this case, server 1 is used for task 1 because it has the smallest weight (4), and after task completion, it gets assigned additional tasks as servers become free.
Approach: This approach simulates the task assignment process by tracking server availability and using a priority queue for efficient task allocation based on server weight and index.
Observations:
• Servers should be processed in order of their availability and weight.
• Use a priority queue to efficiently assign tasks to the smallest available server.
• The problem requires efficient simulation with a focus on managing availability and task queueing. A priority queue is ideal for maintaining the smallest available server.
Steps:
• Create a priority queue to store free servers with their weights and indices.
• For each task, assign it to the server with the smallest weight and index.
• Update the server status after a task is assigned and track when the server becomes free.
Empty Inputs:
• If there are no servers or no tasks, the result should be an empty array.
Large Inputs:
• Make sure the algorithm can handle large numbers of servers and tasks efficiently.
Special Values:
• If servers or tasks contain extreme values, ensure proper handling of edge cases.
Constraints:
• The approach should work efficiently with the given constraints (up to 2 * 10^5 servers and tasks).
vector<int> assignTasks(vector<int>& servers, vector<int>& tasks) {
priority_queue<array<long, 3>, vector<array<long, 3>>, greater<array<long, 3>>> avail, busy;
int n = servers.size();
for(int i = 0; i < n; i++)
avail.push({0, servers[i], i});
int m = tasks.size();
vector<int> ans;
for(int t = 0; t < m; t++) {
while(!busy.empty() && (busy.top()[0] <= t || avail.empty())) {
auto [time, w, idx] = busy.top();
busy.pop();
avail.push({time <= t? 0: time, w, idx});
}
auto [time, w, idx] = avail.top();
avail.pop();
busy.push({ max(time, (long)t) + tasks[t], w, idx});
ans.push_back(idx);
}
return ans;
}
1 : Initial Setup
vector<int> assignTasks(vector<int>& servers, vector<int>& tasks) {
Defines the function `assignTasks` that takes in two vectors: `servers` representing the server capacities and `tasks` representing the tasks to be completed.
2 : Queue Initialization
priority_queue<array<long, 3>, vector<array<long, 3>>, greater<array<long, 3>>> avail, busy;
Declares two priority queues: `avail` for tracking available servers and `busy` for tracking servers that are currently occupied with tasks.
3 : Server Assignment
int n = servers.size();
Gets the number of servers by finding the size of the `servers` vector.
4 : Queue Population
for(int i = 0; i < n; i++)
Loops through each server and pushes a tuple containing the initial time (0), the server's capacity, and the server index into the `avail` priority queue.
5 : Task Counting
avail.push({0, servers[i], i});
Pushes each server into the `avail` queue with the initial time set to 0.
6 : Task Size Setup
int m = tasks.size();
Stores the number of tasks by finding the size of the `tasks` vector.
7 : Answer Initialization
vector<int> ans;
Initializes a vector `ans` to store the indices of the servers assigned to each task.
8 : Task Assignment Loop
for(int t = 0; t < m; t++) {
Loops through each task in the `tasks` vector.
9 : Server Availability Check
while(!busy.empty() && (busy.top()[0] <= t || avail.empty())) {
Checks if any server is free or if there are available servers that need to be assigned tasks.
10 : Pop Busy Server
auto [time, w, idx] = busy.top();
Pops the top of the `busy` queue to get the earliest available server.
11 : Push to Available Queue
busy.pop();
Removes the server from the `busy` queue as it becomes free.
12 : Reassign Available Server
avail.push({time <= t? 0: time, w, idx});
Pushes the server back into the `avail` queue with updated time.
13 : Assign Task to Server
auto [time, w, idx] = avail.top();
Assigns the task to the server at the top of the `avail` queue.
14 : Update Available Queue
avail.pop();
Removes the assigned server from the `avail` queue.
15 : Server Occupy
busy.push({ max(time, (long)t) + tasks[t], w, idx});
Pushes the server into the `busy` queue with the updated time when it will be free.
16 : Log Task Assignment
ans.push_back(idx);
Adds the server index to the `ans` vector, indicating that the task has been assigned to this server.
17 : Return Result
return ans;
Returns the vector `ans` containing the indices of the servers to which tasks were assigned.
Best Case: O(m log n) where m is the number of tasks and n is the number of servers.
Average Case: O(m log n)
Worst Case: O(m log n)
Description: In the worst case, each task involves an insertion and extraction from the priority queue, leading to a time complexity of O(m log n).
Best Case: O(n + m)
Worst Case: O(n + m) due to the space required for the priority queues and the task assignments.
Description: The space complexity is dominated by the need to store the server states and task assignments.
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