LeetCode 1834: Single-Threaded CPU Solution

Master LeetCode problem 1834 (Single-Threaded CPU), a medium 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.

1834. Single-Threaded CPU

Medium

Array Sorting Heap (Priority Queue)

Problem Explanation

Explanation

To solve this problem, we need to simulate the CPU processing tasks based on the given rules. We can use a priority queue to keep track of the available tasks sorted by their processing times. We iterate through the tasks based on their enqueue times and process them accordingly.

Create a priority queue to store tasks based on their processing times and indices.
Sort the tasks based on their enqueue times.
Iterate through the tasks:
- Enqueue tasks that are available at the current time into the priority queue.
- If the CPU is idle and there are available tasks in the queue, process the task with the shortest processing time.
- Update the current time and continue until all tasks are processed.

Time Complexity: O(n log n) where n is the number of tasks. Space Complexity: O(n) for the priority queue.

Solution Code

import java.util.*;

class Solution {
    public int[] getOrder(int[][] tasks) {
        int n = tasks.length;
        int[] order = new int[n];
        PriorityQueue<int[]> pq = new PriorityQueue<>((a, b) -> a[1] == b[1] ? a[0] - b[0] : a[1] - b[1]);
        Queue<Integer> availableTasks = new LinkedList<>();
        
        for (int i = 0; i < n; i++) {
            tasks[i] = new int[] {i, tasks[i][0], tasks[i][1]}; // {index, enqueueTime, processingTime}
        }
        Arrays.sort(tasks, (a, b) -> a[1] - b[1]);
        
        int time = tasks[0][1];
        int idx = 0;
        
        while (idx < n || !pq.isEmpty()) {
            while (idx < n && tasks[idx][1] <= time) {
                pq.offer(new int[] {tasks[idx][0], tasks[idx][2]});
                idx++;
            }
            
            if (pq.isEmpty()) {
                time = tasks[idx][1];
                continue;
            }
            
            int[] task = pq.poll();
            order[idx] = task[0];
            time += task[1];
            availableTasks.add(task[0]);
        }
        
        return order;
    }
}

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