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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; 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 } }

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