What is the time complexity of Task Scheduler?

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

What data structures are used to solve Task Scheduler?

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

What are the key insights for solving Task Scheduler?

Insight 1: The task with the highest frequency dictates the minimum length of the schedule. We need to focus on the most frequent tasks first. Insight 2: The schedule can be visualized as 'slots' where the most frequent tasks are placed, separated by the cooling interval `n`. These slots may need to be padded with idle intervals.

Which companies ask LeetCode 621. Task Scheduler in interviews?

This problem is frequently asked at DE Shaw, DoorDash, IBM, Intuit, Remitly and 6 more companies.

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Task Scheduler - LeetCode 621 Solution

Task Scheduler - Complete Solution Guide

Task Scheduler is LeetCode problem 621, 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 an array of CPU tasks , each labeled with a letter from A to Z, and a number n . Each CPU interval can be idle or allow the completion of one task. Tasks can be completed in any order, but there's a constraint: there has to be a gap of at least n intervals between two tasks with the same label. Return the minimum number of CPU intervals required to complete all tasks. Example 1: Input: tasks = ["A","A","A","B","B","B"], n = 2 Output: 8 Explanation: A possible sequence is: A -> B ->

Detailed Explanation

The Task Scheduler problem requires you to determine the minimum number of CPU intervals needed to execute a given list of tasks. Each task is represented by an uppercase letter. There's a cooling period `n` between executing the same task again. You can execute tasks in any order, and the CPU can be idle during an interval if no task is available for execution. The goal is to find the shortest possible schedule that completes all tasks while respecting the cooling constraint.

Solution Approach

The solution calculates the minimum length based on the frequency of the most frequent task. First, it counts the occurrences of each task. Then, it identifies the maximum frequency and the number of tasks with that maximum frequency. The minimum length is calculated using the formula: `(max_freq - 1) * (n + 1) + num_tasks_with_max_freq`. Finally, the algorithm returns the maximum of the calculated minimum length and the total number of tasks, ensuring all tasks are scheduled.

Step-by-Step Algorithm

Step 1: Count the frequency of each task using a hash map (Counter in Python, HashMap in Java/C++, manual counting in C).
Step 2: Find the maximum frequency among all tasks.
Step 3: Count the number of tasks that have the maximum frequency.
Step 4: Calculate the minimum length of the schedule using the formula: `(max_freq - 1) * (n + 1) + num_tasks_with_max_freq`.
Step 5: Return the maximum of the calculated minimum length and the total number of tasks. This ensures that if there are many different tasks with low frequencies, the schedule will be long enough to accommodate them without needing idle slots.

Key Insights

Insight 1: The task with the highest frequency dictates the minimum length of the schedule. We need to focus on the most frequent tasks first.
Insight 2: The schedule can be visualized as 'slots' where the most frequent tasks are placed, separated by the cooling interval `n`. These slots may need to be padded with idle intervals.
Insight 3: If the number of tasks is greater than the calculated minimum length based on the most frequent task, then the total number of tasks determines the length.

Complexity Analysis

Time Complexity: O(N)

Space Complexity: O(1)

Topics

This problem involves: Array, Hash Table, Greedy, Sorting, Heap (Priority Queue), Counting.

Companies

Asked at: DE Shaw, DoorDash, IBM, Intuit, Remitly, Roblox, Rubrik, Salesforce, Snowflake, Zeta, zeta suite.

Task Scheduler | LeetCode 621

Medium

Array Hash Table Greedy Sorting Heap (Priority Queue)Counting

Time: O(N)Space: O(1)

Solution Code

Available in:

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

class Solution {
    public int leastInterval(char[] tasks, int n) {
        if (tasks == null || tasks.length == 0) {
            return 0;
        }

        Map<Character, Integer> taskCounts = new HashMap<>();
        for (char task : tasks) {
            taskCounts.put(task, taskCounts.getOrDefault(task, 0) + 1);
        }

        int maxFreq = 0;
        for (char task : taskCounts.keySet()) {
            maxFreq = Math.max(maxFreq, taskCounts.get(task));
        }

        int numTasksWithMaxFreq = 0;
        for (char task : taskCounts.keySet()) {
            if (taskCounts.get(task) == maxFreq) {
                numTasksWithMaxFreq++;
            }
        }

        int minLength = (maxFreq - 1) * (n + 1) + numTasksWithMaxFreq;

        return Math.max(tasks.length, minLength);
    }
}

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Problem Info

Description

Problem 621: Task Scheduler - Detailed explanation and solution approaches.

Topics

Array Hash Table Greedy Sorting Heap (Priority Queue)Counting

Companies

DE ShawDoorDashIBMIntuitRemitlyRobloxRubrikSalesforce

Stats

Difficulty:Medium

Languages:4

Companies:11

Problem #:621

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

5 steps

1
Step 1: Count the frequency of each task using a hash map (Counter in Python, HashMap in Java/C++, manual counting in C).
2
Step 2: Find the maximum frequency among all tasks.
3
Step 3: Count the number of tasks that have the maximum frequency.
4
Step 4: Calculate the minimum length of the schedule using the formula: `(max_freq - 1) * (n + 1) + num_tasks_with_max_freq`.
5
Step 5: Return the maximum of the calculated minimum length and the total number of tasks. This ensures that if there are many different tasks with low frequencies, the schedule will be long enough to accommodate them without needing idle slots.

💡

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)

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is dominated by the linear traversal of the tasks array to count the frequency of each task. The space complexity depends on the data structure used to store the frequencies. Since we are dealing with uppercase English letters, the size of the hash map will be at most 26, making the space complexity constant, O(1).

📖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

Priority Queue (Heap)

Use a priority queue to store the tasks and their frequencies. At each interval, select the task with the highest frequency that is not in its cooling period. Insert idle intervals when no task is available.

Time: O(N log 26) ~ O(N) because we are limited to 26 letters. Each task is added and removed from the priority queue at most once, and each operation on the priority queue takes O(log 26) time.Space: O(26) ~ O(1). The priority queue holds a maximum of 26 distinct tasks.

Trade-offs: This approach is more intuitive and simulates the actual scheduling process but has a slightly higher constant factor due to the heap operations compared to the main solution. The heap based solution may be easier to understand at a high level.

✨

Key Insights

1Insight 1: The task with the highest frequency dictates the minimum length of the schedule. We need to focus on the most frequent tasks first.
2Insight 2: The schedule can be visualized as 'slots' where the most frequent tasks are placed, separated by the cooling interval `n`. These slots may need to be padded with idle intervals.
3Insight 3: If the number of tasks is greater than the calculated minimum length based on the most frequent task, then the total number of tasks determines the length.

📚

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!

Greedy Algorithm

Making the best choice at each step without looking ahead.

Sorting

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

Edge Cases

Edge case 1: Empty input list of tasks. The code handles this by returning 0, as there are no tasks to schedule.
Edge case 2: `n = 0`. This means there is no cooling period, so the minimum length is simply the number of tasks. The formula correctly handles this case because the `(max_freq - 1) * (n + 1)` becomes 0 when n=0. Thus the final result is `max(tasks.length, num_tasks_with_max_freq)` which will resolve to tasks.length.
Edge case 3: All tasks are distinct. In this case, `max_freq` will be 1, and `num_tasks_with_max_freq` will be the number of tasks. The formula becomes `(1-1) * (n+1) + tasks.length`, which simplifies to `tasks.length`.
Edge case 4: All tasks are the same. The formula will correctly schedule all tasks separated by n intervals. For example, tasks = ['A', 'A', 'A'], n=2 gives (3-1)*(2+1) + 1 = 7. Correct execution would be A _ _ A _ _ A.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not considering the case where the number of tasks is greater than the calculated minimum length. For example, if tasks = ['A', 'B', 'C', 'D', 'E', 'F'], n = 2, and all tasks have frequency 1, the formula might calculate a smaller length than the actual 6 tasks.

✗Common mistake 2: Incorrectly calculating the number of idle slots required. The formula `(max_freq - 1) * (n + 1)` gives the number of slots needed to accommodate all but the last instance of each most frequent task, plus the slots needed for the idle intervals. `num_tasks_with_max_freq` then corrects the schedule length by accounting for the last occurrences of the most frequent task.

✅

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 the execution of processes on a CPU, ensuring that processes with high priority or frequent dependencies are given preference while respecting resource constraints and cooldown periods (e.g., preventing overheating).

🏦Real-world scenario 2: Scheduling tasks in a distributed computing system, where certain tasks require a cooldown period before being executed again on the same node. This algorithm can help minimize the overall completion time for the task set.

🎮Real-world scenario 3: Managing resources in a factory or assembly line where different operations have cooling or setup times between them.

💡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

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ 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: August 21, 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

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

@AlgoMasterJul 13

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

@InterviewProJul 6

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJun 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"?

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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