What is the time complexity of Minimum Processing Time?

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

What data structures are used to solve Minimum Processing Time?

This problem uses Array, Greedy, Sorting. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Minimum Processing Time?

Insight 1: Sorting `processorTime` in ascending order and `tasks` in descending order allows us to greedily assign the longest tasks to the earliest available processors. Insight 2: The maximum time for each processor is determined by the processor's availability time plus the longest task assigned to it.

Which companies ask LeetCode 2895. Minimum Processing Time in interviews?

This problem is frequently asked at Akuna Capital, Nutanix.

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Minimum Processing Time - LeetCode 2895 Solution

Minimum Processing Time - Complete Solution Guide

Minimum Processing Time is LeetCode problem 2895, 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 have a certain number of processors, each having 4 cores. The number of tasks to be executed is four times the number of processors. Each task must be assigned to a unique core, and each core can only be used once. You are given an array processorTime representing the time each processor becomes available and an array tasks representing how long each task takes to complete. Return the minimum time needed to complete all tasks. Example 1: Input: processorTime = [8,10], tasks = [2,2,3,1,8,7,4,

Detailed Explanation

The problem requires finding the minimum time needed to complete a set of tasks given a number of processors with 4 cores each. Each task must be assigned to a unique core, and each core can only be used once. We are given `processorTime`, representing the time each processor becomes available, and `tasks`, representing the time it takes to complete each task. The goal is to assign tasks to processors such that the maximum completion time among all processors is minimized. The input `tasks` array will always have a length that is four times the length of the `processorTime` array.

Solution Approach

The solution uses a greedy approach. It sorts the `processorTime` in ascending order and the `tasks` in descending order. Then, it iterates through the processors and assigns the four longest remaining tasks to each processor. For each processor, it calculates the completion time as the processor's availability time plus the time of its longest task. The maximum of these completion times across all processors is the minimum time needed to complete all tasks.

Step-by-Step Algorithm

Step 1: Sort the `processorTime` array in ascending order.
Step 2: Sort the `tasks` array in descending order.
Step 3: Iterate through the sorted `processorTime` array from index 0 to n-1, where n is the number of processors.
Step 4: In each iteration, calculate the completion time for the current processor. This is the sum of the processor's availability time and the task at index `i * 4` in the sorted `tasks` array. Since `tasks` is sorted descending, index `i*4` gives us the largest task assigned to the i-th processor.
Step 5: Update the `max_time` with the maximum completion time encountered so far.
Step 6: Return the `max_time` after iterating through all processors.

Key Insights

Insight 1: Sorting `processorTime` in ascending order and `tasks` in descending order allows us to greedily assign the longest tasks to the earliest available processors.
Insight 2: The maximum time for each processor is determined by the processor's availability time plus the longest task assigned to it.
Insight 3: The overall minimum processing time is the maximum of all processor completion times.

Complexity Analysis

Time Complexity: O(n log n)

Space Complexity: O(1)

Topics

This problem involves: Array, Greedy, Sorting.

Companies

Asked at: Akuna Capital, Nutanix.

Minimum Processing Time | LeetCode 2895

Medium

Array Greedy Sorting

Time: O(n log n)Space: O(1)

Solution Code

Available in:

import java.util.Collections;
import java.util.List;

class Solution {
    public int minProcessingTime(List<Integer> processorTime, List<Integer> tasks) {
        Collections.sort(processorTime);
        Collections.sort(tasks, Collections.reverseOrder());
        
        int n = processorTime.size();
        int max_time = 0;
        
        for (int i = 0; i < n; i++) {
            int current_processor_finish_time = processorTime.get(i) + tasks.get(i * 4);
            if (current_processor_finish_time > max_time) {
                max_time = current_processor_finish_time;
            }
        }
        
        return max_time;
    }
}

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Problem Info

Description

Problem 2895: Minimum Processing Time - Detailed explanation and solution approaches.

Topics

Array Greedy Sorting

Companies

Akuna CapitalNutanix

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:2895

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: Sort the `processorTime` array in ascending order.
2
Step 2: Sort the `tasks` array in descending order.
3
Step 3: Iterate through the sorted `processorTime` array from index 0 to n-1, where n is the number of processors.
4
Step 4: In each iteration, calculate the completion time for the current processor. This is the sum of the processor's availability time and the task at index `i * 4` in the sorted `tasks` array. Since `tasks` is sorted descending, index `i*4` gives us the largest task assigned to the i-th processor.
5
Step 5: Update the `max_time` with the maximum completion time encountered so far.
6
Step 6: Return the `max_time` after iterating through all processors.

💡

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 log n)

✨ Log-linear - Great for sorting algorithms!

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The dominant operations are the sorting of `processorTime` and `tasks`. Most standard sorting algorithms (e.g., quicksort, mergesort, heapsort) have a time complexity of O(n log n), where n is the number of elements being sorted. The rest of the algorithm involves simple arithmetic operations and comparisons within a loop that iterates n times, resulting in O(n) time complexity. Because O(n log n) dominates O(n), the overall time complexity is O(n log n). The space complexity is O(1) because the sorting can be done in place and we are using a constant amount of extra space.

📖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

Using a priority queue

Sort the tasks array in ascending order. Use a priority queue (min-heap) to store the available processors. For each set of 4 tasks, poll the processor with the smallest available time, add the task times to the processor's available time, and push the new available time back into the priority queue. After all tasks are assigned, find the maximum time in the priority queue.

Time: O(n log n) for initial task sorting and O(n log n) for heap operations (n tasks / 4 * log n), so overall O(n log n).Space: O(n) for the priority queue.

Trade-offs: Using a priority queue might be more intuitive for some, but it has a higher space complexity than the greedy approach. The time complexity remains the same, but the constant factors could be higher due to heap operations. It's useful when you want to explicitly manage the processor availability in a queue-like structure.

✨

Key Insights

1Insight 1: Sorting `processorTime` in ascending order and `tasks` in descending order allows us to greedily assign the longest tasks to the earliest available processors.
2Insight 2: The maximum time for each processor is determined by the processor's availability time plus the longest task assigned to it.
3Insight 3: The overall minimum processing time is the maximum of all processor completion times.

📚

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: Small input arrays (e.g., `processorTime` with only one element). The algorithm correctly handles this case as the loop will execute only once, and the correct maximum completion time will be calculated.
Edge case 2: `processorTime` elements are equal. The algorithm works fine because `processorTime` is sorted in ascending order but this has no impact since the assignment of tasks is independent of processorTime order (after sorting).
Edge case 3: All tasks have the same execution time. Algorithm works because `tasks` is sorted in descending order, but this has no impact since all tasks have the same execution time.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not sorting the arrays properly. Sorting `processorTime` in descending order or `tasks` in ascending order will not lead to the optimal solution.

✗Common mistake 2: Incorrectly calculating the completion time for each processor. Forgetting to consider the longest task assigned to each processor will result in an incorrect answer.

✗Common mistake 3: Not considering the maximum completion time across all processors. The final result should be the maximum of all processor completion times, not the sum or average.

✅

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 in a multi-core processor system. The algorithm can be used to efficiently assign tasks to processor cores to minimize the overall completion time.

🏦Real-world scenario 2: Resource allocation in a cloud computing environment. Cloud providers can use this algorithm to allocate virtual machines (processors) to user tasks to optimize resource utilization and minimize execution time.

💡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.

👨‍💻 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 16, 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

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

@InterviewProJun 30

❓

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.

What are common greedy problem patterns?

Activity selection (sort by end time), interval scheduling (sort by start/end), Huffman coding (always combine smallest), fractional knapsack (sort by value/weight ratio), and minimum spanning tree. Look for problems where local decisions don't affect future options.

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 "Minimum Processing Time"?

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

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