The Employee That Worked on the Longest Task - Practice Coding | SlaveCode
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2432. The Employee That Worked on the Longest Task
Easy
10 Points
Array
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:
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.
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.
2432. The Employee That Worked on the Longest Task
Easy
10 Points
Array
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:
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.
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.