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The Employee That Worked on the Longest Task

EasyUpdated: Aug 2, 2025
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Problem

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 thesmallest id among them.

Examples

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.

Solution

Method 1 - One Pass

We compute the duration of each task and keep track of the maximum duration and the smallest employee id who worked on it.

Code

C++
#include <vector>
using namespace std;
int hardestWorker(int n, vector<vector<int>>& logs) {
    int maxTime = logs[0][1], res = logs[0][0];
    for (int i = 1; i < logs.size(); ++i) {
        int t = logs[i][1] - logs[i-1][1];
        if (t > maxTime || (t == maxTime && logs[i][0] < res)) {
            maxTime = t;
            res = logs[i][0];
        }
    }
    return res;
}
Java
class Solution {
    public int hardestWorker(int n, int[][] logs) {
        int maxTime = logs[0][1], res = logs[0][0];
        for (int i = 1; i < logs.length; ++i) {
            int t = logs[i][1] - logs[i-1][1];
            if (t > maxTime || (t == maxTime && logs[i][0] < res)) {
                maxTime = t;
                res = logs[i][0];
            }
        }
        return res;
    }
}
Python
from typing import List
def hardestWorker(n: int, logs: List[List[int]]) -> int:
    max_time = logs[0][1]
    res = logs[0][0]
    for i in range(1, len(logs)):
        t = logs[i][1] - logs[i-1][1]
        if t > max_time or (t == max_time and logs[i][0] < res):
            max_time = t
            res = logs[i][0]
    return res
Rust
pub fn hardest_worker(_n: i32, logs: Vec<Vec<i32>>) -> i32 {
    let mut max_time = logs[0][1];
    let mut res = logs[0][0];
    for i in 1..logs.len() {
        let t = logs[i][1] - logs[i-1][1];
        if t > max_time || (t == max_time && logs[i][0] < res) {
            max_time = t;
            res = logs[i][0];
        }
    }
    res
}
TypeScript
function hardestWorker(n: number, logs: number[][]): number {
    let maxTime = logs[0][1], res = logs[0][0];
    for (let i = 1; i < logs.length; ++i) {
        const t = logs[i][1] - logs[i-1][1];
        if (t > maxTime || (t === maxTime && logs[i][0] < res)) {
            maxTime = t;
            res = logs[i][0];
        }
    }
    return res;
}

Complexity

  • ⏰ Time complexity: O(n) where n is the number of logs.
  • 🧺 Space complexity: O(1)

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