# 2432. The Employee That Worked on the Longest Task

## Description

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.

## Solutions

Solution 1: Direct Traversal

We use a variable $last$ to record the end time of the last task, a variable $mx$ to record the longest working time, and a variable $ans$ to record the employee with the longest working time and the smallest $id$. Initially, all three variables are $0$.

Next, we traverse the array $logs$. For each employee, we subtract the end time of the last task from the time the employee completes the task to get the working time $t$ of this employee. If $mx$ is less than $t$, or $mx$ equals $t$ and the $id$ of this employee is less than $ans$, then we update $mx$ and $ans$. Then we update $last$ to be the end time of the last task plus $t$. Continue to traverse until the entire array is traversed.

Finally, return the answer $ans$.

The time complexity is $O(n)$, where $n$ is the length of the array $logs$. The space complexity is $O(1)$.

• class Solution {
public int hardestWorker(int n, int[][] logs) {
int ans = 0;
int last = 0, mx = 0;
for (int[] log : logs) {
int uid = log[0], t = log[1];
t -= last;
if (mx < t || (mx == t && ans > uid)) {
ans = uid;
mx = t;
}
last += t;
}
return ans;
}
}

• class Solution {
public:
int hardestWorker(int n, vector<vector<int>>& logs) {
int ans = 0, mx = 0, last = 0;
for (auto& log : logs) {
int uid = log[0], t = log[1];
t -= last;
if (mx < t || (mx == t && ans > uid)) {
mx = t;
ans = uid;
}
last += t;
}
return ans;
}
};

• class Solution:
def hardestWorker(self, n: int, logs: List[List[int]]) -> int:
last = mx = ans = 0
for uid, t in logs:
t -= last
if mx < t or (mx == t and ans > uid):
ans, mx = uid, t
last += t
return ans


• func hardestWorker(n int, logs [][]int) (ans int) {
var mx, last int
for _, log := range logs {
uid, t := log[0], log[1]
t -= last
if mx < t || (mx == t && uid < ans) {
mx = t
ans = uid
}
last += t
}
return
}

• function hardestWorker(n: number, logs: number[][]): number {
let [ans, mx, last] = [0, 0, 0];
for (let [uid, t] of logs) {
t -= last;
if (mx < t || (mx == t && ans > uid)) {
ans = uid;
mx = t;
}
last += t;
}
return ans;
}


• impl Solution {
pub fn hardest_worker(n: i32, logs: Vec<Vec<i32>>) -> i32 {
let mut res = 0;
let mut max = 0;
let mut pre = 0;
let t = log[1] - pre;
if t > max || (t == max && res > log[0]) {
res = log[0];
max = t;
}
pre = log[1];
}
res
}
}