LeetCode in Kotlin
1665. Minimum Initial Energy to Finish Tasks
Hard
You are given an array tasks where tasks[i] = [actuali, minimumi]:
actualiis the actual amount of energy you spend to finish theithtask.minimumiis the minimum amount of energy you require to begin theithtask.
For example, if the task is [10, 12] and your current energy is 11, you cannot start this task. However, if your current energy is 13, you can complete this task, and your energy will be 3 after finishing it.
You can finish the tasks in any order you like.
Return the minimum initial amount of energy you will need to finish all the tasks.
Example 1:
Input: tasks = [[1,2],[2,4],[4,8]]
Output: 8
Explanation:
Starting with 8 energy, we finish the tasks in the following order:
-
3rd task. Now energy = 8 - 4 = 4.
-
2nd task. Now energy = 4 - 2 = 2.
-
1st task. Now energy = 2 - 1 = 1.
Notice that even though we have leftover energy, starting with 7 energy does not work because we cannot do the 3rd task.
Example 2:
Input: tasks = [[1,3],[2,4],[10,11],[10,12],[8,9]]
Output: 32
Explanation:
Starting with 32 energy, we finish the tasks in the following order:
-
1st task. Now energy = 32 - 1 = 31.
-
2nd task. Now energy = 31 - 2 = 29.
-
3rd task. Now energy = 29 - 10 = 19.
-
4th task. Now energy = 19 - 10 = 9.
-
5th task. Now energy = 9 - 8 = 1.
Example 3:
Input: tasks = [[1,7],[2,8],[3,9],[4,10],[5,11],[6,12]]
Output: 27
Explanation:
Starting with 27 energy, we finish the tasks in the following order:
-
5th task. Now energy = 27 - 5 = 22.
-
2nd task. Now energy = 22 - 2 = 20.
-
3rd task. Now energy = 20 - 3 = 17.
-
1st task. Now energy = 17 - 1 = 16.
-
4th task. Now energy = 16 - 4 = 12.
-
6th task. Now energy = 12 - 6 = 6.
Constraints:
1 <= tasks.length <= 1051 <= actuali <= minimumi <= 104
Solution
class Solution {
fun minimumEffort(tasks: Array<IntArray>): Int {
tasks.sortWith { a: IntArray, b: IntArray -> a[1] - a[0] - b[1] + b[0] }
var prev = 0
for (item in tasks) {
prev = Math.max(prev + item[0], item[1])
}
return prev
}
}