What is the time complexity of Parallel Courses III?

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

What is the space complexity of Parallel Courses III?

The space complexity is O(n + m).

What data structures are used to solve Parallel Courses III?

This problem uses Array, Dynamic Programming, Graph, Topological Sort. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Parallel Courses III?

Insight 1: The problem can be modeled as a directed acyclic graph (DAG), where courses are nodes and prerequisites are edges. The topological order of the graph represents a valid sequence in which the courses can be taken. Insight 2: Dynamic programming (DP) can be used to compute the earliest completion time for each course. The completion time of a course is the maximum of the completion times of its prerequisites plus its own duration.

Which companies ask LeetCode 2050. Parallel Courses III in interviews?

This problem is frequently asked at Acko, Citadel, Snap, Snowflake, Stripe and 2 more companies.

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Parallel Courses III - LeetCode 2050 Solution

Parallel Courses III - Complete Solution Guide

Parallel Courses III is LeetCode problem 2050, 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

You are given an integer n , which indicates that there are n courses labeled from 1 to n . You are also given a 2D integer array relations where relations[j] = [prevCourse j , nextCourse j ] denotes that course prevCourse j has to be completed before course nextCourse j (prerequisite relationship). Furthermore, you are given a 0-indexed integer array time where time[i] denotes how many months it takes to complete the (i+1) th course. You must find the minimum number of months needed to complete

Detailed Explanation

The problem asks us to find the minimum number of months required to complete 'n' courses. Each course has a specific duration (given in the 'time' array). Some courses have prerequisites, defined by the 'relations' array, meaning a course can only be started after its prerequisites are completed. We can take multiple courses simultaneously. The goal is to determine the earliest possible completion time for all courses, considering dependencies and course durations.

Solution Approach

The solution uses a combination of topological sort and dynamic programming. First, the input relations are used to build an adjacency list representing the course dependencies. The in-degree of each course is also calculated. Courses with an in-degree of 0 (no prerequisites) are added to a queue. Then, a topological sort is performed using Kahn's algorithm. While processing the queue, we update the completion time of each course's neighbors using dynamic programming. Finally, the maximum completion time among all courses is returned, which represents the minimum time to complete all courses.

Step-by-Step Algorithm

Step 1: Build the adjacency list (graph) from the 'relations' input and compute the in-degree of each course.
Step 2: Initialize a 'completion_time' array to store the earliest completion time for each course. Initialize a queue with all courses that have an in-degree of 0 (no prerequisites). Set their completion time to their respective duration from the 'time' array.
Step 3: Perform topological sort using Kahn's algorithm. While the queue is not empty, dequeue a course 'u'.
Step 4: For each neighbor 'v' of 'u' (course 'v' depends on 'u'), update the completion time of 'v' as max(current completion time of 'v', completion time of 'u' + time[v-1]).
Step 5: Decrement the in-degree of 'v'. If the in-degree of 'v' becomes 0, enqueue 'v'.
Step 6: After the topological sort is complete, find the maximum value in the 'completion_time' array. This is the minimum time required to complete all courses.

Key Insights

Insight 1: The problem can be modeled as a directed acyclic graph (DAG), where courses are nodes and prerequisites are edges. The topological order of the graph represents a valid sequence in which the courses can be taken.
Insight 2: Dynamic programming (DP) can be used to compute the earliest completion time for each course. The completion time of a course is the maximum of the completion times of its prerequisites plus its own duration.
Insight 3: Topological sorting is crucial for processing the courses in a valid order, ensuring that prerequisites are completed before their dependent courses.

Complexity Analysis

Time Complexity: O(n + m)

Space Complexity: O(n + m)

Topics

This problem involves: Array, Dynamic Programming, Graph, Topological Sort.

Companies

Asked at: Acko, Citadel, Snap, Snowflake, Stripe, TikTok, Two Sigma.

Parallel Courses III | LeetCode 2050

Hard

Array Dynamic Programming Graph Topological Sort

Time: O(n + m)Space: O(n + m)

Solution Code

Available in:

import java.util.*;

class Solution {
    public int minimumTime(int n, int[][] relations, int[] time) {
        Map<Integer, List<Integer>> adj = new HashMap<>();
        int[] inDegree = new int[n + 1];

        for (int[] relation : relations) {
            int u = relation[0];
            int v = relation[1];
            adj.computeIfAbsent(u, k -> new ArrayList<>()).add(v);
            inDegree[v] += 1;
        }

        int[] completionTime = new int[n + 1];
        Queue<Integer> queue = new LinkedList<>();

        for (int i = 1; i <= n; i++) {
            if (inDegree[i] == 0) {
                queue.offer(i);
                completionTime[i] = time[i - 1];
            }
        }

        while (!queue.isEmpty()) {
            int u = queue.poll();

            if (adj.containsKey(u)) {
                for (int v : adj.get(u)) {
                    completionTime[v] = Math.max(completionTime[v], completionTime[u] + time[v - 1]);
                    inDegree[v] -= 1;

                    if (inDegree[v] == 0) {
                        queue.offer(v);
                    }
                }
            }
        }

        int maxCompletionTime = 0;
        for (int i = 1; i <= n; i++) {
            maxCompletionTime = Math.max(maxCompletionTime, completionTime[i]);
        }

        return maxCompletionTime;
    }
}

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Problem Info

Description

Problem 2050: Parallel Courses III - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming Graph Topological Sort

Companies

AckoCitadelSnapSnowflakeStripeTikTokTwo Sigma

Stats

Difficulty:Hard

Languages:4

Companies:7

Problem #:2050

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

6 steps

1
Step 1: Build the adjacency list (graph) from the 'relations' input and compute the in-degree of each course.
2
Step 2: Initialize a 'completion_time' array to store the earliest completion time for each course. Initialize a queue with all courses that have an in-degree of 0 (no prerequisites). Set their completion time to their respective duration from the 'time' array.
3
Step 3: Perform topological sort using Kahn's algorithm. While the queue is not empty, dequeue a course 'u'.
4
Step 4: For each neighbor 'v' of 'u' (course 'v' depends on 'u'), update the completion time of 'v' as max(current completion time of 'v', completion time of 'u' + time[v-1]).
5
Step 5: Decrement the in-degree of 'v'. If the in-degree of 'v' becomes 0, enqueue 'v'.
6
Step 6: After the topological sort is complete, find the maximum value in the 'completion_time' array. This is the minimum time required to complete all courses.

💡

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 + m)

💾

Space Complexity

O(n + m)

📝

Why?

The time complexity is determined by the graph traversal during the topological sort. The space complexity is dominated by the storage of the adjacency list representing the graph and the auxiliary arrays used for in-degree, completion time, and the queue.

📖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) with memoization

Instead of topological sort with Kahn's algorithm, DFS can be used to traverse the graph. Memoization is essential to avoid recomputing the completion time for already visited nodes. The completion time of a node can be determined by recursively calculating the completion times of its prerequisites.

Time: O(n + m)Space: O(n + m)

Trade-offs: DFS with memoization offers a similar performance to the topological sort approach. However, it might be slightly less intuitive for some developers. It uses recursion and might consume more stack space in some cases, depending on the graph's structure.

✨

Key Insights

1Insight 1: The problem can be modeled as a directed acyclic graph (DAG), where courses are nodes and prerequisites are edges. The topological order of the graph represents a valid sequence in which the courses can be taken.
2Insight 2: Dynamic programming (DP) can be used to compute the earliest completion time for each course. The completion time of a course is the maximum of the completion times of its prerequisites plus its own duration.
3Insight 3: Topological sorting is crucial for processing the courses in a valid order, ensuring that prerequisites are completed before their dependent courses.

📚

Key Terms Explained

For Students

Array

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

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Graph

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

Edge Cases

Edge case 1: No relations exist (all courses can be taken in parallel). The algorithm correctly handles this by adding all courses to the initial queue and setting their completion times. The maximum of the individual course durations is then returned.
Edge case 2: A single long chain of prerequisites. The algorithm correctly computes the cumulative time required to complete the chain by propagating the completion times through the topological sort.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to initialize the completion time of courses with no prerequisites. This can lead to incorrect calculation of completion times for dependent courses.

✗Common mistake 2: Not using topological sort or dynamic programming and trying to iterate through relations directly without considering dependencies properly. This will likely result in an incorrect answer or potentially infinite loop for more complex graph structures.

✗Common mistake 3: Not finding the MAX completion time at the end. One must take the max value of the courses' completion times to get the overall time.

✅

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 with dependencies. For example, in a software project, tasks have dependencies, and each task requires a specific amount of time. This algorithm can be used to determine the minimum time required to complete the entire project.

🏦Real-world scenario 2: Course scheduling in a university. Courses have prerequisites, and students can take multiple courses simultaneously. This algorithm can be used to determine the minimum semester time required to complete a set of courses.

💡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

It's okay to not finish hard problems in interviews. Show your problem-solving process.

⚠️ 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 10, 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 14

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

@AlgoMasterJul 7

❓

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 identify if a problem needs dynamic programming?

DP problems typically have: 1) Overlapping subproblems (same calculation repeated), 2) Optimal substructure (optimal solution uses optimal solutions to subproblems). Look for keywords like "minimum/maximum," "count ways," "is it possible," or "longest/shortest."

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 "Parallel Courses 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 dynamic programming techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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