Single Threaded CPU Leetcode Problem 1834 [Python Solution]

In this blog post, we will explore the Single Threaded CPU LeetCode problem, specifically problem number 1834.

We’ll break down the Single Threaded CPU LeetCode problem, discuss constraints, provide a brute-force approach, and then dive into an efficient Python solution.

By the end, you’ll have a clear understanding of how to tackle this problem.

Problem Overview

Problem Statement:

You are given n tasks labeled from 0 to n – 1 represented by a 2D integer array tasks, where tasks[i] = [enqueueTimei, processingTimei] means that the i-th task will be available to process at enqueueTimei and will take processingTimei to finish processing.

You have a single-threaded CPU that can process at most one task at a time and will act as follows:

1. If the CPU is idle and there are no available tasks to process, the CPU remains idle.
2. If the CPU is idle and there are available tasks, the CPU will choose the one with the shortest processing time.

If multiple tasks have the same shortest processing time, it will choose the task with the smallest index.

3. Once a task is started, the CPU will process the entire task without stopping.
4. The CPU can finish a task and then start a new one instantly.

Return the order in which the CPU will process the tasks.

Example 1:

Input: tasks = [[1,2],[2,4],[3,2],[4,1]]
Output: [0,2,3,1]

Example 2:

Input: tasks = [[7,10],[7,12],[7,5],[7,4],[7,2]]
Output: [4,3,2,0,1]

Constraints:

• tasks.length == n
• 1 <= n <= 105
• 1 <= enqueueTimei, processingTimei <= 109

Understanding the Constraints

Before we dive into the solution, let’s understand the constraints given in the problem:

• The number of tasks, n, can be as large as 105, which means we need an efficient solution to handle a large input size.
• The enqueue time and processing time for each task can range from 1 to 109. This means the values can be quite large, so our solution should be able to handle such large numbers.
• The tasks are represented as a 2D integer array, where each task has an enqueue time and a processing time.

We need to process these tasks in a specific order.

Now that we have a good understanding of the problem and its constraints, let’s proceed to the solution.

Brute Force Approach

A brute-force approach to solving this problem would involve simulating the CPU’s behavior step by step and determining which task to process at each time step.

This approach would involve:

1. Sorting the tasks by their enqueue time.
2. Iterating through time, tracking which tasks are available to be processed at each time step.
3. Choosing the task with the shortest processing time at each step.
4. Processing the chosen task and recording its index.
5. Repeating the process until all tasks are processed.

However, this approach can be inefficient for large input sizes, as it involves multiple iterations and sorting operations.

Instead, we can use a more efficient approach using a priority queue (min heap) to optimize the solution.

Efficient Approach with Priority Queue

To solve this problem efficiently, we will use a priority queue (min heap) to keep track of the available tasks at any given time.

We’ll also maintain a variable to represent the current time and another variable to store the result, which will be the order in which tasks are processed.

Here’s the Python code for the efficient solution:

getOrder(self, tasks: List[List[int]]) -> List[int]:
    tasks = sorted([(t[0], t[1], i) for i, t in enumerate(tasks)])
    result, heap = [], []
    cur_task_index = 0
    cur_time = tasks[0][0]

    while len(result) < len(tasks):
        while (cur_task_index < len(tasks)) and (tasks[cur_task_index][0] <= cur_time):
            heapq.heappush(heap, (tasks[cur_task_index][1], tasks[cur_task_index][2]))
            cur_task_index += 1
        if heap:
            time_difference, original_index = heapq.heappop(heap)
            cur_time += time_difference
            result.append(original_index)
        elif cur_task_index < len(tasks):
            cur_time = tasks[cur_task_index][0]

    return result

Let’s break down how this efficient approach works step by step:

1. We start by sorting the tasks based on their enqueue time.

We also add the original index of each task to the sorted list to keep track of their order.

2. We initialize two lists: result to store the order of processed tasks and heap as our priority queue (min heap).

We also initialize cur_task_index to keep track of the current task index and cur_time to represent the current time.

3. We enter a loop that continues until we have processed all tasks.

Inside the loop:

• We add tasks to the heap that are currently available for processing.

These are the tasks with enqueue times less than or equal to the current time.

We use a while loop to add all available tasks.

• We pop tasks from the heap to process.

The task with the shortest processing time will be at the top of the min heap.

We record the time difference (processing time) and the original index of the task.

• We update the cur_time by adding the time difference.

This represents the time at which the task is completed.

• We add the original index of the task to the result list to record the order in which tasks are processed.
• If the heap is empty, it means there are no available tasks to process at the moment.

In this case, we set cur_time to the enqueue time of the next task in the sorted list.

4. Finally, we return the result list, which contains the order in which tasks were processed.

This efficient approach minimizes the number of operations required to process the tasks and is capable of handling large input sizes efficiently.

Time and Space Complexity

Time Complexity:

The time complexity of this solution is dominated by the sorting of the tasks, which takes O(n*log(n)) time, where ‘n’ is the number of tasks.

The subsequent processing of tasks using the min heap has a time complexity of O(n*log(n)) as well.

Therefore, the overall time complexity is O(n*log(n)).

Space Complexity:

The space complexity of this solution is O(n) for the result list, O(n) for the heap, and O(n) for the sorted tasks list.

Therefore, the overall space complexity is O(n).

Reasoning Behind Our Approach

The reasoning behind this approach lies in efficiently simulating the behavior of the single-threaded CPU using a priority queue (min heap).

By sorting the tasks based on their enqueue time, we can process them in the correct order while minimizing the number of operations needed to choose the next task to process.

The

use of a min heap allows us to always select the task with the shortest processing time, ensuring that we follow the problem’s criteria.

This approach is both efficient and intuitive, making it a suitable solution for the Single Threaded Cpu problem.

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Conclusion

In this blog post, we tackled the Single Threaded Cpu LeetCode problem (problem number 1834) and provided an efficient Python solution.

We discussed the problem’s details, constraints, and provided both a brute-force approach and a more optimized solution using a priority queue (min heap).

The priority queue-based solution allows us to efficiently process tasks in the correct order, meeting the criteria specified in the problem statement.

By understanding the problem, its constraints, and reasoning through the solution, we can approach similar problems with confidence.

If you found this post helpful or have any questions, suggestions, or comments, please feel free to engage with us.

Don’t forget to check out the problem on LeetCode and practice your skills.

Your support and feedback are greatly appreciated!

Happy coding!