What is the time complexity of Single-Threaded CPU?

The optimal solution has a time complexity of O(n log n).

What data structures are used to solve Single-Threaded CPU?

This problem uses Array, Sorting, Heap (Priority Queue). Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Single-Threaded CPU?

Insight 1: We need to keep track of the current time and which tasks are available to be processed at that time. Insight 2: A min-heap (priority queue) is ideal for selecting the task with the shortest processing time from the available tasks.

Which companies ask LeetCode 1834. Single-Threaded CPU in interviews?

This problem is frequently asked at DoorDash, IBM.

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Single-Threaded CPU - LeetCode 1834 Solution

Single-Threaded CPU - Complete Solution Guide

Single-Threaded CPU is LeetCode problem 1834, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You are given n tasks labeled from 0 to n - 1 represented by a 2D integer array tasks , where tasks[i] = [enqueueTime i , processingTime i ] means that the i th task will be available to process at enqueueTime i and will take processingTime i to finish processing. You have a single-threaded CPU that can process at most one task at a time and will act in the following way: If the CPU is idle and there are no available tasks to process, the CPU remains idle. If the CPU is idle an

Detailed Explanation

The problem asks us to simulate a single-threaded CPU processing a set of tasks. Each task has an `enqueueTime` (when it becomes available) and a `processingTime` (how long it takes to complete). The CPU chooses tasks based on the shortest processing time first, and if there's a tie, it chooses the task with the smaller index. We need to return the order in which the CPU processes the tasks.

Solution Approach

The solution involves sorting the tasks by their enqueue time and then using a min-heap to manage available tasks. The algorithm iterates, simulating the CPU's behavior. It adds tasks to the heap as they become available and then selects the task with the shortest processing time from the heap. The current time is advanced by the processing time of the selected task, and the process continues until all tasks have been processed.

Step-by-Step Algorithm

Step 1: Create a list of tuples/arrays where each tuple contains the enqueue time, processing time, and original index of the task. This preserves the original index for the final result.
Step 2: Sort the list of tasks based on their enqueue times.
Step 3: Initialize an empty min-heap (priority queue) to store available tasks, sorted by processing time (and index as a tiebreaker).
Step 4: Initialize an empty list to store the result (the order of processed tasks).
Step 5: Initialize `current_time` to 0 and `task_idx` to 0. `current_time` represents the CPU's current time, and `task_idx` points to the next task in the sorted list of tasks.
Step 6: While the result list has fewer than `n` elements (where `n` is the total number of tasks), repeat steps 7-9:
Step 7: Add tasks to the min-heap if their enqueue time is less than or equal to the current time. Increment `task_idx` accordingly.
Step 8: If the min-heap is not empty, extract the task with the shortest processing time. Add its index to the result list, and update the current time by adding the task's processing time.
Step 9: If the min-heap is empty, but there are still tasks waiting to be enqueued, advance the current time to the enqueue time of the next task. This prevents the CPU from idling unnecessarily.
Step 10: Return the result list.

Key Insights

Insight 1: We need to keep track of the current time and which tasks are available to be processed at that time.
Insight 2: A min-heap (priority queue) is ideal for selecting the task with the shortest processing time from the available tasks.
Insight 3: Sorting the tasks by their enqueue time allows us to efficiently determine when tasks become available.

Complexity Analysis

Time Complexity: O(n log n)

Space Complexity: O(n)

Topics

This problem involves: Array, Sorting, Heap (Priority Queue).

Companies

Asked at: DoorDash, IBM.

Single-Threaded CPU | LeetCode 1834

Medium

Array Sorting Heap (Priority Queue)

Time: O(n log n)Space: O(n)

Solution Code

Available in:

import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
import java.util.PriorityQueue;

class Solution {
    public int[] getOrder(int[][] tasks) {
        int n = tasks.length;
        int[][] indexedTasks = new int[n][3];
        for (int i = 0; i < n; i++) {
            indexedTasks[i][0] = tasks[i][0];
            indexedTasks[i][1] = tasks[i][1];
            indexedTasks[i][2] = i;
        }

        Arrays.sort(indexedTasks, (a, b) -> Integer.compare(a[0], b[0]));

        List<Integer> resultList = new ArrayList<>();
        PriorityQueue<int[]> availableTasksHeap = new PriorityQueue<>((a, b) -> {
            if (a[0] != b[0]) {
                return Integer.compare(a[0], b[0]);
            } else {
                return Integer.compare(a[1], b[1]);
            }
        });

        int taskIdx = 0;
        long currentTime = 0;

        while (resultList.size() < n) {
            while (taskIdx < n && indexedTasks[taskIdx][0] <= currentTime) {
                int processingTime = indexedTasks[taskIdx][1];
                int originalIndex = indexedTasks[taskIdx][2];
                availableTasksHeap.offer(new int[]{processingTime, originalIndex});
                taskIdx++;
            }

            if (!availableTasksHeap.isEmpty()) {
                int[] currentTask = availableTasksHeap.poll();
                int processingTime = currentTask[0];
                int originalIndex = currentTask[1];
                resultList.add(originalIndex);
                currentTime += processingTime;
            } else if (taskIdx < n) {
                currentTime = indexedTasks[taskIdx][0];
            }
        }

        int[] result = new int[n];
        for (int i = 0; i < n; i++) {
            result[i] = resultList.get(i);
        }

        return result;
    }
}

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Problem Info

Description

Problem 1834: Single-Threaded CPU - Detailed explanation and solution approaches.

Topics

Array Sorting Heap (Priority Queue)

Companies

DoorDashIBM

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:1834

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

10 steps

1
Step 1: Create a list of tuples/arrays where each tuple contains the enqueue time, processing time, and original index of the task. This preserves the original index for the final result.
2
Step 2: Sort the list of tasks based on their enqueue times.
3
Step 3: Initialize an empty min-heap (priority queue) to store available tasks, sorted by processing time (and index as a tiebreaker).
4
Step 4: Initialize an empty list to store the result (the order of processed tasks).
5
Step 5: Initialize `current_time` to 0 and `task_idx` to 0. `current_time` represents the CPU's current time, and `task_idx` points to the next task in the sorted list of tasks.
6
Step 6: While the result list has fewer than `n` elements (where `n` is the total number of tasks), repeat steps 7-9:
7
Step 7: Add tasks to the min-heap if their enqueue time is less than or equal to the current time. Increment `task_idx` accordingly.
8
Step 8: If the min-heap is not empty, extract the task with the shortest processing time. Add its index to the result list, and update the current time by adding the task's processing time.
9
Step 9: If the min-heap is empty, but there are still tasks waiting to be enqueued, advance the current time to the enqueue time of the next task. This prevents the CPU from idling unnecessarily.
10
Step 10: Return the result list.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n log n)

✨ Log-linear - Great for sorting algorithms!

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is dominated by the initial sorting of the tasks, which takes O(n log n) time. The heap operations (insertion and extraction) each take O(log n) time, and these operations are performed at most n times. Therefore, the overall time complexity is O(n log n). The space complexity is O(n) because we store all the tasks in the indexed_tasks array and potentially in the heap.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

No Sorting

Instead of pre-sorting, we can iterate through the original `tasks` array at each time step to find available tasks. This avoids the initial sorting step.

Time: O(n^2 log n)Space: O(n)

Trade-offs: This approach avoids the initial O(n log n) sorting but introduces an O(n) search for available tasks at each time step. Since the heap operations are also performed within the loop, the overall complexity becomes O(n^2 log n). This is less efficient than the sorted approach when `n` is large. Useful if modifying original task is not permitted or for some specific input constraints.

✨

Key Insights

1Insight 1: We need to keep track of the current time and which tasks are available to be processed at that time.
2Insight 2: A min-heap (priority queue) is ideal for selecting the task with the shortest processing time from the available tasks.
3Insight 3: Sorting the tasks by their enqueue time allows us to efficiently determine when tasks become available.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: All tasks have the same enqueue time. The solution correctly handles this by adding all tasks to the heap and then processing them based on processing time (and index if there's a tie).
Edge case 2: The CPU is idle and no tasks are available. The algorithm correctly advances the current time to the next available task, avoiding an infinite loop.
Edge case 3: Large range of enqueue and processing times. The algorithm works correctly even with large values for enqueueTime and processingTime, as it uses long type for currentTime in Java and C++ solutions to avoid integer overflow.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to include the original index of the tasks, which is needed to resolve ties in processing time.

✗Common mistake 2: Not using a long long (or equivalent) to store the current time. The current time can become very large, potentially leading to integer overflow.

✗Common mistake 3: Not handling the case where the CPU is idle and no tasks are available. This can lead to an infinite loop.

✗Common mistake 4: Incorrectly comparing processing times in the heap. Make sure that you are comparing processing times first before comparing indices. Ensure stability in comparison operation when dealing with priority queue.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Task scheduling in operating systems. Operating systems need to schedule tasks on the CPU based on priority or resource requirements.

🏦Real-world scenario 2: Job scheduling in distributed systems. Distributed systems need to schedule jobs across multiple machines, often considering factors like resource availability and job dependencies.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: July 5, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 13

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 6

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Single-Threaded CPU"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether sorting techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?