LeetCode in Kotlin
2589. Minimum Time to Complete All Tasks
Hard
There is a computer that can run an unlimited number of tasks at the same time. You are given a 2D integer array tasks where tasks[i] = [starti, endi, durationi] indicates that the ith task should run for a total of durationi seconds (not necessarily continuous) within the inclusive time range [starti, endi].
You may turn on the computer only when it needs to run a task. You can also turn it off if it is idle.
Return the minimum time during which the computer should be turned on to complete all tasks.
Example 1:
Input: tasks = [[2,3,1],[4,5,1],[1,5,2]]
Output: 2
Explanation:
- The first task can be run in the inclusive time range [2, 2].
- The second task can be run in the inclusive time range [5, 5].
- The third task can be run in the two inclusive time ranges [2, 2] and [5, 5]. The computer will be on for a total of 2 seconds.
Example 2:
Input: tasks = [[1,3,2],[2,5,3],[5,6,2]]
Output: 4
Explanation:
- The first task can be run in the inclusive time range [2, 3].
- The second task can be run in the inclusive time ranges [2, 3] and [5, 5].
- The third task can be run in the two inclusive time range [5, 6].
The computer will be on for a total of 4 seconds.
Constraints:
1 <= tasks.length <= 2000tasks[i].length == 31 <= starti, endi <= 20001 <= durationi <= endi - starti + 1
Solution
class Solution {
fun findMinimumTime(tasks: Array<IntArray>): Int {
var res = 0
val arr = BooleanArray(2001)
tasks.sortWith { a: IntArray, b: IntArray ->
a[1] - b[1]
}
for (task in tasks) {
val start = task[0]
val end = task[1]
val dur = task[2]
var cur = 0
for (i in start..end) {
if (arr[i]) {
cur++
}
}
if (cur < dur) {
var i = end
while (i >= start && cur < dur) {
if (!arr[i]) {
arr[i] = true
res++
cur++
}
i--
}
}
}
return res
}
}