LeetCode in Kotlin

3683. Earliest Time to Finish One Task

Easy

You are given a 2D integer array tasks where tasks[i] = [si, ti].

Each [si, ti] in tasks represents a task with start time si that takes ti units of time to finish.

Return the earliest time at which at least one task is finished.

Example 1:

Input: tasks = [[1,6],[2,3]]

Output: 5

Explanation:

The first task starts at time t = 1 and finishes at time 1 + 6 = 7. The second task finishes at time 2 + 3 = 5. You can finish one task at time 5.

Example 2:

Input: tasks = [[100,100],[100,100],[100,100]]

Output: 200

Explanation:

All three tasks finish at time 100 + 100 = 200.

Constraints:

• 1 <= tasks.length <= 100
• tasks[i] = [si, ti]
• 1 <= si, ti <= 100

Solution

import kotlin.math.min

class Solution {
    fun earliestTime(tasks: Array<IntArray>): Int {
        var ans = 1000
        for (i in tasks.indices) {
            val st = tasks[i][0]
            val tm = tasks[i][1]
            ans = min(ans, st + tm)
        }
        return ans
    }
}

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