LeetCode 3156: Employee Task Duration and Concurrent Tasks Solution

Master LeetCode problem 3156 (Employee Task Duration and Concurrent Tasks), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

3156. Employee Task Duration and Concurrent Tasks

Hard

Database

Problem Explanation

Explanation:

To solve this problem, we can use a priority queue to keep track of the tasks assigned to employees. We iterate through the tasks and assign them to employees based on their availability. If an employee is busy with a task, we check if the current task can be assigned to them based on the start time of the task and the end time of the current task the employee is working on. We also need to keep track of the maximum number of concurrent tasks at any given time.

Algorithm:

Sort the tasks based on their start times.
Initialize a priority queue to store the end times of tasks currently assigned to employees.
Iterate through the sorted tasks:
- If the priority queue is empty, assign the task to an employee and add the end time of the task to the priority queue.
- If the priority queue is not empty, check if the start time of the current task is greater than or equal to the end time of the task at the top of the priority queue. If so, assign the task to an employee and update the end time in the priority queue. Otherwise, assign the task to a new employee and update the maximum concurrent tasks count.
Return the maximum concurrent tasks count.

Time Complexity: O(n log n) (sorting the tasks) Space Complexity: O(n)

Solution Code

import java.util.*;

class Task {
    int start;
    int end;

    public Task(int start, int end) {
        this.start = start;
        this.end = end;
    }
}

class Solution {
    public int maxConcurrentTasks(List<Task> tasks) {
        if (tasks == null || tasks.size() == 0) {
            return 0;
        }

        Collections.sort(tasks, (a, b) -> a.start - b.start);

        PriorityQueue<Integer> pq = new PriorityQueue<>();
        int maxConcurrent = 0;

        for (Task task : tasks) {
            if (!pq.isEmpty() && pq.peek() <= task.start) {
                pq.poll();
            } else {
                maxConcurrent++;
            }
            pq.offer(task.end);
        }

        return maxConcurrent;
    }
}

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