What is the time complexity of Maximum Number of Tasks You Can Assign?

The optimal solution has a time complexity of O(n*log(n)*log(min(n,m))).

What data structures are used to solve Maximum Number of Tasks You Can Assign?

This problem uses Array, Two Pointers, Binary Search, Greedy, Queue, Sorting, Monotonic Queue. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Maximum Number of Tasks You Can Assign?

Insight 1: Sorting both tasks and workers allows us to efficiently use a greedy approach within a binary search framework. Insight 2: Binary search is applicable because the problem asks for the *maximum* number of tasks. We can check if a given number of tasks is possible to complete, allowing binary search on the possible number of tasks.

Which companies ask LeetCode 2071. Maximum Number of Tasks You Can Assign in interviews?

This problem is frequently asked at Coupang, IBM, Walmart Labs.

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Maximum Number of Tasks You Can Assign - LeetCode 2071 Solution

Maximum Number of Tasks You Can Assign - Complete Solution Guide

Maximum Number of Tasks You Can Assign is LeetCode problem 2071, 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 have n tasks and m workers. Each task has a strength requirement stored in a 0-indexed integer array tasks , with the i th task requiring tasks[i] strength to complete. The strength of each worker is stored in a 0-indexed integer array workers , with the j th worker having workers[j] strength. Each worker can only be assigned to a single task and must have a strength greater than or equal to the task's strength requirement (i.e., workers[j] >= tasks[i] ). Additionally, you have pills magical

Detailed Explanation

The problem asks us to find the maximum number of tasks that can be completed given a set of tasks with strength requirements, a set of workers with strength, a number of pills, and the strength increase each pill provides. Each worker can only be assigned to one task, and their strength must be greater than or equal to the task's strength requirement. Pills can be given to workers to increase their strength, but each worker can receive at most one pill. The goal is to maximize the number of assigned tasks.

Solution Approach

The solution uses a binary search approach to find the maximum number of tasks that can be completed. For each possible number of tasks (the 'mid' value in the binary search), a helper function `can_assign` checks if it's possible to assign that many tasks. The `can_assign` function sorts the first k tasks (easiest k tasks) and workers, then iterates through the tasks in reverse order of strength requirement. For each task, it checks if there are workers strong enough to complete it. If not, it attempts to use a pill on a worker. If no worker can complete the task even with a pill, the function returns `false`. The binary search adjusts the search space based on the result of `can_assign` to find the optimal number of tasks.

Step-by-Step Algorithm

Step 1: Sort the `tasks` and `workers` arrays in ascending order.
Step 2: Perform a binary search on the possible number of tasks, from 0 to the minimum of the lengths of `tasks` and `workers`.
Step 3: Inside the binary search, define a helper function `can_assign(k)` that takes an integer `k` as input, representing the number of tasks we want to try to assign.
Step 4: In `can_assign(k)`, initialize `pills_remaining` to the initial number of `pills`. Also, initialize a deque `overqualified_workers` to store workers that are strong enough to complete the task without a pill. Initialize the `worker_ptr` pointing to the strongest worker (m-1).
Step 5: Iterate through the `k` easiest tasks in reverse order of difficulty (from hardest to easiest).
Step 6: For each task, find all workers strong enough without a pill (add them to the `overqualified_workers` deque from strongest to weakest).
Step 7: If there are overqualified workers for a given task, assign the weakest available worker (pop from the end of the deque).
Step 8: If no overqualified worker available, check if we can use a pill to make an underqualified worker strong enough for the task. Pick the strongest underqualified worker (workers[worker_ptr]), and if even with a pill that worker can't complete the task, then assigning this many tasks (k) is impossible, return `false`.
Step 9: If there are no overqualified workers and no pills available, then assigning this many tasks (k) is impossible, return `false`.
Step 10: If `can_assign(k)` returns `true`, it means we can assign at least `k` tasks, so update the `ans` and search for a larger number of tasks (update `low` to `mid + 1`).
Step 11: If `can_assign(k)` returns `false`, it means we cannot assign `k` tasks, so search for a smaller number of tasks (update `high` to `mid - 1`).
Step 12: After the binary search completes, return the maximum number of tasks `ans` that could be assigned.

Key Insights

Insight 1: Sorting both tasks and workers allows us to efficiently use a greedy approach within a binary search framework.
Insight 2: Binary search is applicable because the problem asks for the *maximum* number of tasks. We can check if a given number of tasks is possible to complete, allowing binary search on the possible number of tasks.
Insight 3: Using a deque to maintain workers who are already strong enough for a task allows us to prioritize using those workers before resorting to pills. This helps conserve pills for later, harder tasks.

Complexity Analysis

Time Complexity: O(n*log(n)*log(min(n,m)))

Space Complexity: O(1)

Topics

This problem involves: Array, Two Pointers, Binary Search, Greedy, Queue, Sorting, Monotonic Queue.

Companies

Asked at: Coupang, IBM, Walmart Labs.

Maximum Number of Tasks You Can Assign | LeetCode 2071

Hard

Array Two Pointers Binary Search Greedy Queue Sorting Monotonic Queue

Time: O(n*log(n)*log(min(n,m)))Space: O(1)

Solution Code

Available in:

import java.util.Arrays;
import java.util.Deque;
import java.util.LinkedList;

class Solution {
    public int maxTaskAssign(int[] tasks, int[] workers, int pills, int strength) {
        int n = tasks.length;
        int m = workers.length;
        Arrays.sort(tasks);
        Arrays.sort(workers);

        // Function to check if k tasks can be assigned
        function can_assign = (k) -> {
            if (k == 0) {
                return true;
            }
            if (k > m) {
                return false;
            }

            int p = pills;
            Deque<Integer> overqualified_workers = new LinkedList<>();
            int worker_ptr = m - 1;

            for (int i = k - 1; i >= 0; i--) {
                int task_strength = tasks[i];

                while (worker_ptr >= 0 && workers[worker_ptr] >= task_strength) {
                    overqualified_workers.add(workers[worker_ptr]);
                    worker_ptr--;
                }

                if (!overqualified_workers.isEmpty()) {
                    overqualified_workers.removeLast();
                } else if (p > 0) {
                    if (worker_ptr >= 0 && workers[worker_ptr] + strength >= task_strength) {
                        p--;
                        worker_ptr--;
                    } else {
                        return false;
                    }
                } else {
                    return false;
                }
            }

            return true;
        };

        int low = 0, high = Math.min(n, m);
        int ans = 0;
        while (low <= high) {
            int mid = (low + high) / 2;
            if (can_assign.apply(mid)) {
                ans = mid;
                low = mid + 1;
            } else {
                high = mid - 1;
            }
        }

        return ans;
    }

    interface function {
        boolean apply(int x);
    }
}

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Problem Info

Description

Problem 2071: Maximum Number of Tasks You Can Assign - Detailed explanation and solution approaches.

Topics

Array Two Pointers Binary Search Greedy Queue Sorting Monotonic Queue

Companies

CoupangIBMWalmart Labs

Stats

Difficulty:Hard

Languages:4

Companies:3

Problem #:2071

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

12 steps

1
Step 1: Sort the `tasks` and `workers` arrays in ascending order.
2
Step 2: Perform a binary search on the possible number of tasks, from 0 to the minimum of the lengths of `tasks` and `workers`.
3
Step 3: Inside the binary search, define a helper function `can_assign(k)` that takes an integer `k` as input, representing the number of tasks we want to try to assign.
4
Step 4: In `can_assign(k)`, initialize `pills_remaining` to the initial number of `pills`. Also, initialize a deque `overqualified_workers` to store workers that are strong enough to complete the task without a pill. Initialize the `worker_ptr` pointing to the strongest worker (m-1).
5
Step 5: Iterate through the `k` easiest tasks in reverse order of difficulty (from hardest to easiest).
6
Step 6: For each task, find all workers strong enough without a pill (add them to the `overqualified_workers` deque from strongest to weakest).
7
Step 7: If there are overqualified workers for a given task, assign the weakest available worker (pop from the end of the deque).
8
Step 8: If no overqualified worker available, check if we can use a pill to make an underqualified worker strong enough for the task. Pick the strongest underqualified worker (workers[worker_ptr]), and if even with a pill that worker can't complete the task, then assigning this many tasks (k) is impossible, return `false`.
9
Step 9: If there are no overqualified workers and no pills available, then assigning this many tasks (k) is impossible, return `false`.
10
Step 10: If `can_assign(k)` returns `true`, it means we can assign at least `k` tasks, so update the `ans` and search for a larger number of tasks (update `low` to `mid + 1`).
11
Step 11: If `can_assign(k)` returns `false`, it means we cannot assign `k` tasks, so search for a smaller number of tasks (update `high` to `mid - 1`).
12
Step 12: After the binary search completes, return the maximum number of tasks `ans` that could be assigned.

💡

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(min(n,m)))

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is dominated by the sorting and the binary search. The sorting is O(n log n) and O(m log m), where n is the number of tasks and m is the number of workers. The binary search has O(log(min(n, m))) iterations. Inside the binary search, the `can_assign` function has a loop that iterates `k` times. The worst case scenario of the `can_assign` function depends on when to use the pills and gather the `overqualified_workers`, but in terms of time complexity, that is the most important part.

📖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) Approach

Instead of sorting workers initially, you can maintain a priority queue (min-heap) of available workers. For each task, you check if the strongest worker can handle it. If not, you give a pill to the weakest suitable worker.

Time: O(n log m), where n is the number of tasks and m is the number of workers. This comes from sorting tasks, and potentially inserting and removing workers from the priority queue.Space: O(m) to store workers in priority queue.

Trade-offs: This approach may be more efficient if the number of workers is significantly larger than the number of tasks. However, implementation complexity is higher.

✨

Key Insights

1Insight 1: Sorting both tasks and workers allows us to efficiently use a greedy approach within a binary search framework.
2Insight 2: Binary search is applicable because the problem asks for the *maximum* number of tasks. We can check if a given number of tasks is possible to complete, allowing binary search on the possible number of tasks.
3Insight 3: Using a deque to maintain workers who are already strong enough for a task allows us to prioritize using those workers before resorting to pills. This helps conserve pills for later, harder tasks.

📚

Key Terms Explained

For Students

Array

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

Two Pointers

Using two indices to traverse data from different positions simultaneously.

Binary Search

Halving the search space each step. Like finding a word in a dictionary!

Queue

FIFO structure - First In, First Out. Like a line at a store!

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: When there are no tasks or workers. The binary search handles this case gracefully because the `low` and `high` variables are initialized correctly, and the while loop condition `low <= high` handles the case where the arrays are empty.
Edge case 2: When the number of pills is 0. The `can_assign` function correctly handles this scenario by skipping the pill-assigning logic, and relying only on worker strength.
Edge case 3: When the strength increase from pills is very large. The algorithm correctly considers the increased strength when evaluating if a worker can complete a task.
Edge case 4: When k = 0. The `can_assign` function will always return `true`, which is the correct behaviour because it means it's always possible to assign zero tasks.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not sorting the tasks and workers. Sorting is crucial for the greedy approach within the `can_assign` function to work correctly.

✗Common mistake 2: Incorrectly implementing the binary search. Ensuring the `low` and `high` pointers are updated correctly and the loop condition is appropriate is vital for finding the optimal solution.

✗Common mistake 3: Not handling edge cases such as empty input arrays or zero pills, resulting in incorrect or runtime errors.

✗Common mistake 4: Not considering the pill usage optimally. The greedy strategy of prioritising the strongest workers first before attempting to use a pill on a weaker worker to complete a task is key to maximizing the solution.

✅

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 in cloud computing. Tasks can be assigned to virtual machines (workers) with different computing power. Pills could represent temporarily increasing VM resources to complete more tasks.

🏦Real-world scenario 2: Scheduling jobs on processors with varying speeds. Some processors might be 'boosted' (given a pill) to complete more demanding jobs.

💡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.

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 31, 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 avoid infinite loops with two pointers?

Always ensure your loop condition guarantees termination (typically left < right or slow != fast). Move at least one pointer in each iteration, and handle edge cases where pointers might not move (like skipping duplicates). Draw out pointer movement on paper first.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

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 "Maximum Number of Tasks You Can Assign"?

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

🤖 Need help with this problem?