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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 theith
task, and 
leaveTimei
is the time at which the employee finished theith
task. All the valuesleaveTimei
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; for log in logs.iter() { 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).