1665. Minimum Initial Energy to Finish Tasks

Description

You are given an array tasks where tasks[i] = [actuali, minimumi]:

• actuali is the actual amount of energy you spend to finish the ith task.
• minimumi is the minimum amount of energy you require to begin the ith task.

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 <= 105
• 1 <= actual​i <= minimumi <= 104

Solutions

• class Solution {
Arrays.sort(tasks, (a, b) -> a[0] - b[0] - (a[1] - b[1]));
int ans = 0, cur = 0;
if (cur < m) {
ans += m - cur;
cur = m;
}
cur -= a;
}
return ans;
}
}

• class Solution {
public:
sort(tasks.begin(), tasks.end(), [&](const auto& a, const auto& b) { return a[0] - a[1] < b[0] - b[1]; });
int ans = 0, cur = 0;
if (cur < m) {
ans += m - cur;
cur = m;
}
cur -= a;
}
return ans;
}
};

• class Solution:
def minimumEffort(self, tasks: List[List[int]]) -> int:
ans = cur = 0
for a, m in sorted(tasks, key=lambda x: x[0] - x[1]):
if cur < m:
ans += m - cur
cur = m
cur -= a
return ans


• func minimumEffort(tasks [][]int) (ans int) {
cur := 0
if cur < m {
ans += m - cur
cur = m
}
cur -= a
}
return
}

• function minimumEffort(tasks: number[][]): number {
tasks.sort((a, b) => a[0] - a[1] - (b[0] - b[1]));
let ans = 0;
let cur = 0;
for (const [a, m] of tasks) {
if (cur < m) {
ans += m - cur;
cur = m;
}
cur -= a;
}
return ans;
}