What is the time complexity of The Employee That Worked on the Longest Task?

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

What data structures are used to solve The Employee That Worked on the Longest Task?

This problem uses Array. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving The Employee That Worked on the Longest Task?

The duration of a task is the difference between its `leaveTime` and the `leaveTime` of the previous task (or 0 for the first task). We need to track both the maximum task duration and the ID of the employee who worked that task. The smallest ID should be prioritized in case of ties.

Which companies ask LeetCode 2432. The Employee That Worked on the Longest Task in interviews?

This problem is frequently asked at IBM, Morgan Stanley.

The problem asks to find the ID of the employee who worked on the longest task. The input is a 2D integer array `logs` where each inner array represents a task. `logs[i] = [id i , leaveTime i ]` indicates employee `id i ` finished task `i` at time `leaveTime i `. Tasks are sequential; task `i` starts immediately after task `i-1` ends (task 0 starts at time 0). The output is the ID of the employee who worked the longest; if there's a tie, return the smallest ID. The problem emphasizes that `leaveTime i ` values are unique and strictly increasing.

Solution in python3. Solution in java. Solution in cpp. Solution in c

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The Employee That Worked on the Longest Task - LeetCode 2432 Solution

The Employee That Worked on the Longest Task - Complete Solution Guide

The Employee That Worked on the Longest Task is LeetCode problem 2432, a Easy 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 employees, each with a unique id from 0 to n - 1 . You are given a 2D integer array logs where logs[i] = [id i , leaveTime i ] where: id i is the id of the employee that worked on the i th task, and leaveTime i is the time at which the employee finished the i th task. All the values leaveTime i are unique . Note that the i th task starts the moment right after the (i - 1) th task ends, and the 0 th task starts at time 0 . Return the id of the employee that worked the task with the lo

Detailed Explanation

The problem asks to find the ID of the employee who worked on the longest task. The input is a 2D integer array `logs` where each inner array represents a task. `logs[i] = [idi, leaveTimei]` indicates employee `idi` finished task `i` at time `leaveTimei`. Tasks are sequential; task `i` starts immediately after task `i-1` ends (task 0 starts at time 0). The output is the ID of the employee who worked the longest; if there's a tie, return the smallest ID. The problem emphasizes that `leaveTimei` values are unique and strictly increasing.

Solution Approach

The solution iterates through the `logs` array. For each task, it calculates the task duration. It keeps track of the maximum duration encountered so far and the ID of the employee who worked that task. If a longer task is found, the maximum duration and employee ID are updated. If a task has the same duration as the current maximum, the employee ID is updated to the minimum of the current ID and the employee ID associated with the maximum duration. This ensures that the smallest employee ID is returned in case of ties.

Step-by-Step Algorithm

Initialize `max_time` to the `leaveTime` of the first task and `employee_id` to the ID of the first task.
Initialize `start_time` to 0.
Iterate through the `logs` array:
Calculate `current_time` as the difference between the current task's `leaveTime` and `start_time`.
If `current_time` is greater than `max_time`, update `max_time` and `employee_id`.
If `current_time` is equal to `max_time`, update `employee_id` to the minimum of `employee_id` and the current task's ID.
Update `start_time` to the current task's `leaveTime`.
Return `employee_id`.

Key Insights

The duration of a task is the difference between its `leaveTime` and the `leaveTime` of the previous task (or 0 for the first task).
We need to track both the maximum task duration and the ID of the employee who worked that task. The smallest ID should be prioritized in case of ties.
Iterating through the `logs` array once is sufficient to find the solution.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Array.

Companies

Asked at: IBM, Morgan Stanley.

The Employee That Worked on the Longest Task | LeetCode 2432

Easy

Array

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

Solution Code

Available in:

class Solution {
    public int hardestWorker(int n, int[][] logs) {
        int longestTime = logs[0][1];
        int hardestWorkerId = logs[0][0];
        int prevLeaveTime = logs[0][1];

        for (int i = 1; i < logs.length; i++) {
            int currentTime = logs[i][1] - prevLeaveTime;
            if (currentTime > longestTime) {
                longestTime = currentTime;
                hardestWorkerId = logs[i][0];
            } else if (currentTime == longestTime) {
                hardestWorkerId = Math.min(hardestWorkerId, logs[i][0]);
            }
            prevLeaveTime = logs[i][1];
        }

        if(longestTime < logs[0][1]){
            longestTime = logs[0][1];
            hardestWorkerId = logs[0][0];
        }

        return hardestWorkerId;
    }
}

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Problem Info

Description

Problem 2432: The Employee That Worked on the Longest Task - Detailed explanation and solution approaches.

Topics

Array

Companies

IBMMorgan Stanley

Stats

Difficulty:Easy

Languages:4

Companies:2

Problem #:2432

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
Initialize `max_time` to the `leaveTime` of the first task and `employee_id` to the ID of the first task.
2
Initialize `start_time` to 0.
3
Iterate through the `logs` array:
4
Calculate `current_time` as the difference between the current task's `leaveTime` and `start_time`.
5
If `current_time` is greater than `max_time`, update `max_time` and `employee_id`.
6
If `current_time` is equal to `max_time`, update `employee_id` to the minimum of `employee_id` and the current task's ID.
7
Update `start_time` to the current task's `leaveTime`.
8
Return `employee_id`.

💡

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)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The loop iterates linearly through the input array, performing constant-time operations within each iteration. No additional data structures that scale with the input size are used.

📖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 HashMap

Create a HashMap to store the total time each employee worked. Iterate through logs, calculating the task duration and adding it to the employee's total time. Finally, find the employee with the maximum total time.

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

Trade-offs: This approach uses more space but might be slightly easier to understand for beginners. The single pass approach is more efficient in space.

✨

Key Insights

1The duration of a task is the difference between its `leaveTime` and the `leaveTime` of the previous task (or 0 for the first task).
2We need to track both the maximum task duration and the ID of the employee who worked that task. The smallest ID should be prioritized in case of ties.
3Iterating through the `logs` array once is sufficient to find the solution.

📚

Key Terms Explained

For Students

Array

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

Edge Cases

The first task might be the longest (handled by initializing `max_time` with the first task's `leaveTime`).
Multiple employees might have worked the longest tasks (handled by choosing the minimum ID among those with the longest task duration).

⚠️

Common Mistakes to Avoid

✗Forgetting to handle the case where multiple employees have the longest task duration and not selecting the smallest ID.

✗Incorrectly calculating the task duration by not considering the previous task's end 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?

💼Analyzing employee productivity and identifying top performers based on task completion time.

🏦Optimizing resource allocation by identifying the most utilized resources or employees.

💡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

Don't rush. Explain your thought process clearly. Easy problems are often about communication.

⚠️ 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: September 1, 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

❓

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 should I approach easy-level algorithm problems?

Easy problems build foundational skills. Focus on: 1) Understanding the problem completely before coding, 2) Writing clean, readable code, 3) Handling all edge cases (empty input, single element, etc.), 4) Analyzing time and space complexity. Don't skip complexity analysis even for simple solutions.

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 "The Employee That Worked on the Longest Task"?

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

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