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Formatted question description: https://leetcode.ca/all/2432.html

# 2432. The Employee That Worked on the Longest Task

• Difficulty: Easy.
• Related Topics: .
• Similar Questions: .

## 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** 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 employees 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 (Java, C++, Python)

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

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

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


• func hardestWorker(n int, logs [][]int) int {
ans, mx, last := 0, 0, 0
for _, e := range logs {
uid, t := e[0], e[1]
x := t - last
if mx < x {
mx, ans = x, uid
} else if mx == x && ans > uid {
ans = uid
}
last = t
}
return ans
}

• function hardestWorker(n: number, logs: number[][]): number {
let [ans, max_num] = logs[0];
for (let i = 1; i < logs.length; i++) {
let duration = logs[i][1] - logs[i - 1][1];
let id = logs[i][0];
if (duration > max_num || (duration == max_num && id < ans)) {
ans = id;
max_num = duration;
}
}
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
}
}



Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).