LeetCode in Kotlin

2432. The Employee That Worked on the Longest Task

Easy

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

class Solution {
    fun hardestWorker(logs: Array<IntArray>): Int {
        var max: Int
        var tid: Int
        var temp: Int
        max = 0
        tid = Int.MAX_VALUE
        var i = 0
        while (i < logs.size) {
            temp = logs[i][1]
            if (i > 0) {
                temp -= logs[i - 1][1]
            }
            if (temp > max) {
                max = temp
                tid = logs[i][0]
            } else if (temp == max && tid > logs[i][0]) {
                tid = logs[i][0]
            }
            i++
        }
        return tid
    }
}

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