What is the time complexity of Task Scheduler II?

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

What data structures are used to solve Task Scheduler II?

This problem uses Array, Hash Table, Simulation. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Task Scheduler II?

Insight 1: A hash table is essential for efficiently tracking the last completion day for each task type, allowing us to quickly determine the earliest possible day to execute the same task again. Insight 2: The key is to simulate the process day by day, checking if a task of the same type has been recently completed and, if so, adjusting the current day accordingly. The 'current_day' variable acts as the simulator's clock.

Which companies ask LeetCode 2365. Task Scheduler II in interviews?

This problem is frequently asked at DoorDash, Duolingo, Nvidia, Remitly.

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Task Scheduler II - LeetCode 2365 Solution

Task Scheduler II - Complete Solution Guide

Task Scheduler II is LeetCode problem 2365, 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 a 0-indexed array of positive integers tasks , representing tasks that need to be completed in order , where tasks[i] represents the type of the i th task. You are also given a positive integer space , which represents the minimum number of days that must pass after the completion of a task before another task of the same type can be performed. Each day, until all tasks have been completed, you must either: Complete the next task from tasks , or Take a break. Return the minimum num

Detailed Explanation

The problem asks us to determine the minimum number of days required to complete a given set of tasks in a specific order. Each task has a type, and after completing a task, we must wait a certain number of days (defined by `space`) before another task of the same type can be performed. We can either complete the next task or take a break on any given day. The goal is to find the optimal scheduling to minimize the total number of days taken.

Solution Approach

The solution uses a greedy approach along with a hash table (or map) to store the last completion day for each task type. The algorithm iterates through the tasks in the given order, simulating the completion process day by day. For each task, it checks if a task of the same type has been completed within the 'space' constraint. If it has, the current day is advanced to the earliest day when the task can be performed. The last completion day of the current task is then updated in the hash table. The final value of the current day represents the minimum number of days needed to complete all tasks.

Step-by-Step Algorithm

Step 1: Initialize a hash table (e.g., a dictionary or map) to store the last completion day for each task type.
Step 2: Initialize a variable 'current_day' to 0. This represents the current day in the simulation.
Step 3: Iterate through the tasks array.
Step 4: For each task, increment the 'current_day' by 1.
Step 5: Check if the current task type exists in the hash table.
Step 6: If the task type exists, retrieve the last completion day and calculate the earliest possible day to perform the task again (last_completion_day + space + 1).
Step 7: Update the 'current_day' to be the maximum of the current 'current_day' and the earliest possible day.
Step 8: Update the last completion day for the current task type in the hash table with the current 'current_day'.
Step 9: After iterating through all tasks, return the final 'current_day'.

Key Insights

Insight 1: A hash table is essential for efficiently tracking the last completion day for each task type, allowing us to quickly determine the earliest possible day to execute the same task again.
Insight 2: The key is to simulate the process day by day, checking if a task of the same type has been recently completed and, if so, adjusting the current day accordingly. The 'current_day' variable acts as the simulator's clock.
Insight 3: The problem can be solved greedily by performing tasks in order. If the same task was recently completed, we fast forward to the earliest possible day.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(u)

Topics

This problem involves: Array, Hash Table, Simulation.

Companies

Asked at: DoorDash, Duolingo, Nvidia, Remitly.

Task Scheduler II | LeetCode 2365

Medium

Array Hash Table Simulation

Time: O(n)Space: O(u)

Solution Code

Available in:

import java.util.HashMap;
import java.util.Map;

class Solution {
    public long taskSchedulerII(int[] tasks, int space) {
        Map<Integer, Long> last_completion_day = new HashMap<>();
        long current_day = 0;
        
        for (int task_type : tasks) {
            current_day += 1;
            if (last_completion_day.containsKey(task_type)) {
                long earliest_possible_day = last_completion_day.get(task_type) + space + 1;
                current_day = Math.max(current_day, earliest_possible_day);
            }
            
            last_completion_day.put(task_type, current_day);
            
        }
        
        return current_day;
    }
}

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Problem Info

Description

Problem 2365: Task Scheduler II - Detailed explanation and solution approaches.

Topics

Array Hash Table Simulation

Companies

DoorDashDuolingoNvidiaRemitly

Stats

Difficulty:Medium

Languages:4

Companies:4

Problem #:2365

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

9 steps

1
Step 1: Initialize a hash table (e.g., a dictionary or map) to store the last completion day for each task type.
2
Step 2: Initialize a variable 'current_day' to 0. This represents the current day in the simulation.
3
Step 3: Iterate through the tasks array.
4
Step 4: For each task, increment the 'current_day' by 1.
5
Step 5: Check if the current task type exists in the hash table.
6
Step 6: If the task type exists, retrieve the last completion day and calculate the earliest possible day to perform the task again (last_completion_day + space + 1).
7
Step 7: Update the 'current_day' to be the maximum of the current 'current_day' and the earliest possible day.
8
Step 8: Update the last completion day for the current task type in the hash table with the current 'current_day'.
9
Step 9: After iterating through all tasks, return the final 'current_day'.

💡

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)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(u)

📝

Why?

The time complexity is determined by the single loop that iterates through the tasks array. Each operation inside the loop (hash table lookups, updates, arithmetic operations) takes constant time on average. The space complexity depends on the number of unique task types, as the hash table stores the last completion day for each unique type. In the worst case, all tasks are of unique types, and the hash table will store n entries. Therefore, the space complexity is O(n) in the worst case and O(u) in general.

📖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

Simulation with array

Simulate the task scheduling with an array representing the days and mark tasks as completed. This can be inefficient for large space values

Time: O(n*space) in the worst case, where n is the number of tasks.Space: O(n*space), depending on how one represents the array and the number of days to be tracked.

Trade-offs: This approach is less efficient because it requires marking off days in an array. It's simpler to understand but doesn't scale well to large 'space' values compared to the hash table approach.

✨

Key Insights

1Insight 1: A hash table is essential for efficiently tracking the last completion day for each task type, allowing us to quickly determine the earliest possible day to execute the same task again.
2Insight 2: The key is to simulate the process day by day, checking if a task of the same type has been recently completed and, if so, adjusting the current day accordingly. The 'current_day' variable acts as the simulator's clock.
3Insight 3: The problem can be solved greedily by performing tasks in order. If the same task was recently completed, we fast forward to the earliest possible day.

📚

Key Terms Explained

For Students

Array

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

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

Edge Cases

Edge case 1: tasks = [1, 2, 3], space = 0. The tasks can be completed in 3 days. The solution handles this correctly as the `current_day` will increment normally.
Edge case 2: tasks = [1, 1, 1, 1], space = 2. The tasks will be completed on days 1, 4, 7, 10. The solution will correctly compute this by advancing the current day accordingly.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly calculating the earliest possible day to perform a task again. Forgetting to add 1 to the `last_completion_day + space` would result in incorrect scheduling.

✗Common mistake 2: Not using a long long to represent the current day in languages such as Java and C++. Given the constaints, current_day could be larger than the maximum value an integer could contain.

✅

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: Optimizing task scheduling in a manufacturing plant where certain machines need a cool-down period after processing specific types of materials. This algorithm can help minimize the overall production time.

🏦Real-world scenario 2: Scheduling maintenance tasks on servers in a data center, where similar servers require a certain downtime between maintenance cycles. This algorithm can ensure efficient resource utilization and minimize downtime across the data center.

💡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.

👨‍💻 From Experience

This problem pattern appears frequently in real codebases. I've used similar logic when building production systems.

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: June 26, 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

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

@DevExpert92Jul 13

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 6

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

How do I choose between a hash map and a hash set?

Use a hash set when you only need to track presence/absence of elements (like detecting duplicates). Use a hash map when you need to associate values with keys (like counting frequencies or storing indices). Hash sets use less memory since they only store keys.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Task Scheduler II"?

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 hash table techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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