What is the time complexity of Course Schedule III?

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

What data structures are used to solve Course Schedule III?

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

What are the key insights for solving Course Schedule III?

Insight 1: Sorting courses by their last day allows us to prioritize courses with earlier deadlines, making it more likely to fit more courses into our schedule. Insight 2: Using a max-heap (priority queue) helps us keep track of the durations of the courses we've currently scheduled. This allows us to efficiently remove the longest course if adding a new course exceeds its deadline.

Which companies ask LeetCode 630. Course Schedule III in interviews?

This problem is frequently asked at Flipkart, Works Applications.

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Course Schedule III - LeetCode 630 Solution

Course Schedule III - Complete Solution Guide

Course Schedule III is LeetCode problem 630, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

There are n different online courses numbered from 1 to n . You are given an array courses where courses[i] = [duration i , lastDay i ] indicate that the i th course should be taken continuously for duration i days and must be finished before or on lastDay i . You will start on the 1 st day and you cannot take two or more courses simultaneously. Return the maximum number of courses that you can take . Example 1: Input: courses = [[100,200],[200,1300],[1000,1250],[2000,3200]] Output: 3 Explanatio

Detailed Explanation

The problem asks us to find the maximum number of courses we can take, given a list of courses. Each course has a duration and a last day it can be completed. We must take each course contiguously, and we cannot take two courses simultaneously. We must start on day 1 and determine the largest number of courses that can be completed within their deadlines.

Solution Approach

The solution uses a greedy approach combined with a max-heap. First, the courses are sorted based on their 'lastDay'. Then, the algorithm iterates through the sorted courses, keeping track of the 'current_time' used and maintaining a max-heap of the course durations that have been added so far. For each course, the duration is added to the current time, and the duration is added to the heap. If the current time exceeds the course's 'lastDay', the largest duration in the heap is removed (which corresponds to the longest course currently scheduled), and the current time is adjusted accordingly. At the end, the number of courses in the heap represents the maximum number of courses that can be taken.

Step-by-Step Algorithm

Step 1: Sort the courses based on their last day (deadline) in ascending order. This prioritizes courses that need to be completed sooner.
Step 2: Initialize a max-heap to store the durations of the courses taken so far. This allows efficient retrieval and removal of the course with the longest duration.
Step 3: Initialize a variable 'current_time' to 0, representing the current time spent taking courses.
Step 4: Iterate through the sorted courses. For each course:
Step 5: Add the course's duration to 'current_time'.
Step 6: Push the course's duration onto the max-heap.
Step 7: If 'current_time' is greater than the course's 'lastDay', it means we've exceeded the deadline. In this case, remove the largest duration from the max-heap and subtract it from 'current_time'. This simulates removing the longest course we've already taken.
Step 8: After iterating through all courses, the size of the max-heap represents the maximum number of courses that can be taken.

Key Insights

Insight 1: Sorting courses by their last day allows us to prioritize courses with earlier deadlines, making it more likely to fit more courses into our schedule.
Insight 2: Using a max-heap (priority queue) helps us keep track of the durations of the courses we've currently scheduled. This allows us to efficiently remove the longest course if adding a new course exceeds its deadline.
Insight 3: The greedy approach of sorting by last day and using a max-heap works because if adding a new course makes us exceed its deadline, we can always remove the longest currently scheduled course to potentially free up enough time to accommodate the new course while still keeping the number of courses taken as high as possible.

Complexity Analysis

Time Complexity: O(nlogn)

Space Complexity: O(n)

Topics

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

Companies

Asked at: Flipkart, Works Applications.

Course Schedule III | LeetCode 630

Hard

Array Greedy Sorting Heap (Priority Queue)

Time: O(nlogn)Space: O(n)

Solution Code

Available in:

import java.util.Arrays;
import java.util.PriorityQueue;

class Solution {
    public int scheduleCourse(int[][] courses) {
        Arrays.sort(courses, (a, b) -> a[1] - b[1]);
        
        PriorityQueue<Integer> max_heap = new PriorityQueue<>((a, b) -> b - a);
        int current_time = 0;
        
        for (int[] course : courses) {
            int duration = course[0];
            int lastDay = course[1];
            
            current_time += duration;
            max_heap.offer(duration);
            
            if (current_time > lastDay) {
                current_time -= max_heap.poll();
            }
        }
        
        return max_heap.size();
    }
}

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Problem Info

Description

Problem 630: Course Schedule III - Detailed explanation and solution approaches.

Topics

Array Greedy Sorting Heap (Priority Queue)

Companies

FlipkartWorks Applications

Stats

Difficulty:Hard

Languages:4

Companies:2

Problem #:630

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

8 steps

1
Step 1: Sort the courses based on their last day (deadline) in ascending order. This prioritizes courses that need to be completed sooner.
2
Step 2: Initialize a max-heap to store the durations of the courses taken so far. This allows efficient retrieval and removal of the course with the longest duration.
3
Step 3: Initialize a variable 'current_time' to 0, representing the current time spent taking courses.
4
Step 4: Iterate through the sorted courses. For each course:
5
Step 5: Add the course's duration to 'current_time'.
6
Step 6: Push the course's duration onto the max-heap.
7
Step 7: If 'current_time' is greater than the course's 'lastDay', it means we've exceeded the deadline. In this case, remove the largest duration from the max-heap and subtract it from 'current_time'. This simulates removing the longest course we've already taken.
8
Step 8: After iterating through all courses, the size of the max-heap represents the maximum number of courses that can be taken.

💡

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(nlogn)

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

Sorting the courses dominates the time complexity. The loop iterates 'n' times, and inside the loop, heap operations (push and pop) take O(log n) time each. Hence, the overall time complexity is O(n log n). The space complexity is determined by the max heap, which can grow up to the size of 'n' if we take all the courses.

📖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

Dynamic Programming

Dynamic Programming could potentially be used, where dp[i][j] represents the maximum number of courses we can take up to course i with time j. However, this becomes less feasible given the constraints of the problem, specifically the upper bound of lastDay being 10^4. This could require a dp table of size n*10^4, leading to memory and potentially time issues.

Time: O(n*T), where T is the largest lastDaySpace: O(n*T)

Trade-offs: Dynamic Programming offers a more structured approach but has higher space complexity and may not be efficient for large deadline values. The greedy approach is generally more efficient in this scenario.

✨

Key Insights

1Insight 1: Sorting courses by their last day allows us to prioritize courses with earlier deadlines, making it more likely to fit more courses into our schedule.
2Insight 2: Using a max-heap (priority queue) helps us keep track of the durations of the courses we've currently scheduled. This allows us to efficiently remove the longest course if adding a new course exceeds its deadline.
3Insight 3: The greedy approach of sorting by last day and using a max-heap works because if adding a new course makes us exceed its deadline, we can always remove the longest currently scheduled course to potentially free up enough time to accommodate the new course while still keeping the number of courses taken as high as possible.

📚

Key Terms Explained

For Students

Array

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

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 (courses = []). The algorithm will correctly return 0 because the loop won't execute, and the initial size of the max-heap will remain 0.
Edge case 2: Courses that cannot be taken (duration > lastDay). These courses will be considered and potentially added to the heap. If taking them results in exceeding deadlines of subsequent courses, the algorithm will remove courses from the heap to compensate, ensuring a valid result.
Edge case 3: Courses with the same last day. The sorting algorithm ensures consistent behavior in such cases, and the subsequent heap operations handle them correctly based on the sorting order.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not sorting the courses by their last day. This leads to a suboptimal solution because we might prioritize longer courses with later deadlines, blocking us from taking more shorter courses with earlier deadlines.

✗Common mistake 2: Using a min-heap instead of a max-heap. We need to be able to quickly identify and remove the longest course we've already taken, which requires a max-heap.

✗Common mistake 3: Incorrectly calculating the 'current_time' when removing a course from the heap. It's crucial to subtract the correct duration when a course is dropped to reflect the accurate time remaining.

✅

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: Resource allocation where tasks have deadlines and different resource consumption amounts. The goal is to maximize the number of tasks completed within their deadlines, given a limited resource capacity (e.g., CPU time, memory).

🏦Real-world scenario 2: Project scheduling, where projects have deadlines and completion times. A project manager can use this approach to determine the maximum number of projects that can be completed by their due dates, given resource constraints.

💡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

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ Watch Out

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

👨‍💻 From Experience

I encountered a variation of this problem in a system design interview. The key was recognizing the underlying pattern.

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 30, 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

❓

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.

How do I verify if a greedy approach is correct?

Prove the greedy choice property: the locally optimal choice leads to globally optimal solution. Use exchange argument: show that swapping any element with the greedy choice doesn't improve the solution. If you can't prove it, test extensively or consider DP.

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

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 "Course Schedule III"?

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

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