Introduction

In this blog post, we will explore how to solve the LeetCode problem "2432" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

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] = [idi, leaveTimei] where: idi is the id of the employee that worked on the ith task, and leaveTimei is the time at which the employee finished the ith task. All the values leaveTimei are unique. Note that the ith task starts the moment right after the (i - 1)th task ends, and the 0th task starts at time 0. Return the id of the employee that worked the task with the longest time. If there is a tie between two or more employees, return the smallest id among them. Example 1: Input: n = 10, logs = [[0,3],[2,5],[0,9],[1,15]] Output: 1 Explanation: Task 0 started at 0 and ended at 3 with 3 units of times. Task 1 started at 3 and ended at 5 with 2 units of times. Task 2 started at 5 and ended at 9 with 4 units of times. Task 3 started at 9 and ended at 15 with 6 units of times. The task with the longest time is task 3 and the employee with id 1 is the one that worked on it, so we return 1. Example 2: Input: n = 26, logs = [[1,1],[3,7],[2,12],[7,17]] Output: 3 Explanation: Task 0 started at 0 and ended at 1 with 1 unit of times. Task 1 started at 1 and ended at 7 with 6 units of times. Task 2 started at 7 and ended at 12 with 5 units of times. Task 3 started at 12 and ended at 17 with 5 units of times. The tasks with the longest time is task 1. The employee that worked on it is 3, so we return 3. Example 3: Input: n = 2, logs = [[0,10],[1,20]] Output: 0 Explanation: Task 0 started at 0 and ended at 10 with 10 units of times. Task 1 started at 10 and ended at 20 with 10 units of times. The tasks with the longest time are tasks 0 and 1. The employees that worked on them are 0 and 1, so we return the smallest id 0. Constraints: 2 <= n <= 500 1 <= logs.length <= 500 logs[i].length == 2 0 <= idi <= n - 1 1 <= leaveTimei <= 500 idi != idi+1 leaveTimei are sorted in a strictly increasing order.

Explanation

Here's a breakdown of the solution:

• Calculate Task Durations: Iterate through the logs array, calculating the duration of each task by subtracting the start time from the end time. The start time for the first task is 0, and for subsequent tasks, it's the end time of the previous task.
• Track Longest Duration: Maintain variables to store the longest duration encountered so far and the corresponding employee ID.

• Handle Ties: If a task duration matches the longest duration, update the employee ID only if the current employee ID is smaller than the previously stored ID.

• Runtime Complexity: O(m), where m is the number of logs.

• Storage Complexity: O(1)

Code

    def hardestWorker(n: int, logs: list[list[int]]) -> int:
    """
    Finds the employee with the longest task time.

    Args:
        n: The number of employees.
        logs: A list of logs, where each log contains the employee ID and leave time.

    Returns:
        The ID of the employee who worked the longest. If there is a tie, the smallest ID is returned.
    """
    longest_time = 0
    employee_id = -1
    start_time = 0

    for log in logs:
        emp_id = log[0]
        leave_time = log[1]
        duration = leave_time - start_time

        if duration > longest_time:
            longest_time = duration
            employee_id = emp_id
        elif duration == longest_time:
            if emp_id < employee_id or employee_id == -1:
                employee_id = emp_id

        start_time = leave_time

    return employee_id