What is the time complexity of Course Schedule?

The optimal solution has a time complexity of O(V+E).

What data structures are used to solve Course Schedule?

This problem uses Depth-First Search, Breadth-First Search, Graph, Topological Sort. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Course Schedule?

Insight 1: The problem can be modeled as a directed graph where courses are nodes and prerequisites are directed edges. Insight 2: Detecting cycles in a directed graph is crucial. Topological sort can be used to detect cycles.

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Course Schedule - LeetCode 207 Solution

Q: What is the space complexity of Course Schedule?

The space complexity is O(V+E).

Q: Which companies ask LeetCode 207. Course Schedule in interviews?

This problem is frequently asked at Anduril, ByteDance, Citadel, Coupang, CrowdStrike and 23 more companies.

Course Schedule - Complete Solution Guide

Course Schedule is LeetCode problem 207, 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

There are a total of numCourses courses you have to take, labeled from 0 to numCourses - 1 . You are given an array prerequisites where prerequisites[i] = [a i , b i ] indicates that you must take course b i first if you want to take course a i . For example, the pair [0, 1] , indicates that to take course 0 you have to first take course 1 . Return true if you can finish all courses. Otherwise, return false . Example 1: Input: numCourses = 2, prerequisites = [[1,0]] Output: true Explanation: The

Detailed Explanation

The problem 'Course Schedule' asks us to determine if it's possible to finish all courses given a number of courses and a set of prerequisites. Each prerequisite is a pair [a, b], meaning that you must take course 'b' before you can take course 'a'. We need to determine if there's a cycle in the dependency graph of courses, because a cycle would make it impossible to complete all courses. If a cycle exists, return false, otherwise return true.

Solution Approach

The provided code uses a topological sort algorithm based on Kahn's algorithm using in-degree. It constructs an adjacency list representation of the course dependencies. It then calculates the in-degree (number of prerequisites) for each course. Courses with an in-degree of 0 are added to a queue. The algorithm then iteratively removes a course from the queue, decrements the in-degree of its dependent courses, and adds dependent courses with an in-degree of 0 to the queue. Finally, it checks if the number of courses finished equals the total number of courses. If they are equal, then all courses can be taken.

Step-by-Step Algorithm

Step 1: Create an adjacency list to represent the course dependencies. `adj[i]` stores a list of courses that depend on course `i`.
Step 2: Calculate the in-degree of each course. The in-degree of a course is the number of prerequisite courses it has.
Step 3: Initialize a queue with courses that have an in-degree of 0 (no prerequisites).
Step 4: While the queue is not empty:
a. Remove a course from the front of the queue.
b. Increment the `courses_finished` counter.
c. For each neighbor (dependent course) of the removed course:
i. Decrement the in-degree of the neighbor.
ii. If the in-degree of the neighbor becomes 0, add it to the queue.
Step 5: Return `true` if `courses_finished` is equal to the total number of courses, indicating that all courses can be taken. Otherwise, return `false`.

Key Insights

Insight 1: The problem can be modeled as a directed graph where courses are nodes and prerequisites are directed edges.
Insight 2: Detecting cycles in a directed graph is crucial. Topological sort can be used to detect cycles.
Insight 3: If a topological sort can be completed successfully (i.e., all nodes are visited), there are no cycles, and all courses can be finished. Otherwise, a cycle exists.

Complexity Analysis

Time Complexity: O(V+E)

Space Complexity: O(V+E)

Topics

This problem involves: Depth-First Search, Breadth-First Search, Graph, Topological Sort.

Companies

Asked at: Anduril, ByteDance, Citadel, Coupang, CrowdStrike, Cruise, DoorDash, Flipkart, Graviton, IXL, Intuit, Karat, LiveRamp, Nordstrom, Nutanix, Nvidia, Oracle, PayPal, Roblox, Snap, Snowflake, Swiggy, VMware, Visa, Walmart Labs, Yelp, Zenefits, eBay.

Course Schedule | LeetCode 207

Medium

Depth-First Search Breadth-First Search Graph Topological Sort

Time: O(V+E)Space: O(V+E)

Solution Code

Available in:

import java.util.ArrayList;
import java.util.LinkedList;
import java.util.List;
import java.util.Queue;

class Solution {
    public boolean canFinish(int numCourses, int[][] prerequisites) {
        List<List<Integer>> adj = new ArrayList<>();
        for (int i = 0; i < numCourses; i++) {
            adj.add(new ArrayList<>());
        }

        int[] in_degree = new int[numCourses];

        for (int[] prerequisite : prerequisites) {
            int course = prerequisite[0];
            int prereq = prerequisite[1];
            adj.get(prereq).add(course);
            in_degree[course]++;
        }

        Queue<Integer> queue = new LinkedList<>();
        for (int i = 0; i < numCourses; i++) {
            if (in_degree[i] == 0) {
                queue.offer(i);
            }
        }

        int courses_finished = 0;
        while (!queue.isEmpty()) {
            int node = queue.poll();
            courses_finished++;

            for (int neighbor : adj.get(node)) {
                in_degree[neighbor]--;
                if (in_degree[neighbor] == 0) {
                    queue.offer(neighbor);
                }
            }
        }

        return courses_finished == numCourses;
    }
}

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Problem Info

Description

Problem 207: Course Schedule - Detailed explanation and solution approaches.

Topics

Depth-First Search Breadth-First Search Graph Topological Sort

Companies

AndurilByteDanceCitadelCoupangCrowdStrikeCruiseDoorDashFlipkart

Stats

Difficulty:Medium

Languages:4

Companies:28

Problem #:207

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 an adjacency list to represent the course dependencies. `adj[i]` stores a list of courses that depend on course `i`.
2
Step 2: Calculate the in-degree of each course. The in-degree of a course is the number of prerequisite courses it has.
3
Step 3: Initialize a queue with courses that have an in-degree of 0 (no prerequisites).
4
Step 4: While the queue is not empty:
5
a. Remove a course from the front of the queue.
6
b. Increment the `courses_finished` counter.
7
c. For each neighbor (dependent course) of the removed course:
8
i. Decrement the in-degree of the neighbor.
9
ii. If the in-degree of the neighbor becomes 0, add it to the queue.
10
Step 5: Return `true` if `courses_finished` is equal to the total number of courses, indicating that all courses can be taken. Otherwise, return `false`.

💡

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(V+E)

💾

Space Complexity

O(V+E)

📝

Why?

The time complexity is determined by the construction of the adjacency list and the topological sort using Kahn's algorithm. The adjacency list construction takes O(E) time. The topological sort involves iterating through all vertices and edges in the worst case, which takes O(V+E) time. Therefore the time complexity is O(V+E). The space complexity comes from the adjacency list which can hold up to E edges and requires O(E) space, and the in-degree array and queue that both store up to V courses each, which require O(V) space. Therefore, the space complexity is O(V+E).

📖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

Depth-First Search (DFS)

Detect cycles using DFS. During DFS, mark nodes as 'visiting' (currently in the recursion stack) and 'visited' (completely explored). If we encounter a 'visiting' node during DFS, it indicates a cycle.

Time: O(V+E)Space: O(V+E) in the worst case due to the recursion stack and adjacency list.

Trade-offs: DFS might be slightly more complex to implement correctly because it requires keeping track of different states of nodes (unvisited, visiting, visited), but it can be slightly more memory efficient in some cases where the maximum depth of the graph is less than the number of vertices.

✨

Key Insights

1Insight 1: The problem can be modeled as a directed graph where courses are nodes and prerequisites are directed edges.
2Insight 2: Detecting cycles in a directed graph is crucial. Topological sort can be used to detect cycles.
3Insight 3: If a topological sort can be completed successfully (i.e., all nodes are visited), there are no cycles, and all courses can be finished. Otherwise, a cycle exists.

📚

Key Terms Explained

For Students

Graph

Nodes connected by edges. Like a social network or city map!

Edge Cases

Edge case 1: No prerequisites (prerequisites list is empty). The algorithm correctly handles this case because the initial in-degree of all courses will be 0, and they will all be added to the queue. `courses_finished` will be equal to `numCourses`, so it returns true.
Edge case 2: A single cycle involving only one course (e.g., [0, 0]). The algorithm will not finish. The in-degree for course 0 will be 1, so it is never added to the queue. `courses_finished` remains 0, so it returns false.
Edge case 3: Multiple disconnected components in the graph. The algorithm correctly processes each component separately since it initializes the queue with all courses that initially have an in-degree of 0.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to handle the case where the prerequisites list is empty. The code should still correctly return true in this case.

✗Common mistake 2: Incorrectly calculating the in-degree of the courses. A wrong in-degree will mess up the topological sort and result in an incorrect answer.

✗Common mistake 3: Not allocating enough space for the adjacency lists or other data structures, leading to potential runtime errors, especially in C.

✅

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. Determining if a set of tasks with dependencies can be completed without circular dependencies.

🏦Real-world scenario 2: Dependency resolution in software packages. Ensuring that all software packages and their dependencies can be installed without creating circular dependencies.

🎮Real-world scenario 3: Compiler design: Ordering of compilation steps to ensure that all dependencies are met.

💡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

Always test your solution with edge cases: empty input, single element, and boundary values.

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 28, 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

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

@InterviewProJul 13

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

@CodeNinjaJul 6

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

@TechLeadJun 29

❓

Frequently Asked Questions

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 "Course Schedule"?

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

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