2432 - The Employee That Worked on the Longest Task (Easy)

Problem Link

https://leetcode.com/problems/the-employee-that-worked-on-the-longest-task/

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, returnthe 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.

Approach 1: Single Pass

This could code be a little bit more compact, but I like unpacking the input and giving it things meaningful names.

• Time complexity: $O(size(logs))$ we need to look at all the entries in logs.
• Space complexity: $O(1)$ only a few integers as state.

C++

Written by @heder

static int hardestWorker(int n, const vector<vector<int>>& logs) {
    int start_time = 0;
    int worker = n;
    int longest_task = 0;
    for (const vector<int>& log : logs) {
        const int id = log[0];
        const int end_time = log[1];
        const int time = end_time - start_time;

        if (time > longest_task || time == longest_task && id < worker) {
            worker = id;
            longest_task = time;
        }

        start_time = end_time;
    }
    return worker;
}