1665. Minimum Initial Energy to Finish Tasks 

Description

You are given an array tasks where tasks[i] = [actual_i, minimum_i]:

actual_i is the actual amount of energy you spend to finish the i^th task.
minimum_i is the minimum amount of energy you require to begin the i^th 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 <= minimum_i <= 10⁴

Solutions

Solution 1: Greedy + Custom Sorting

Assume the number of tasks is $n$ and the initial energy level is $E$. Consider completing the last task. This requires that after completing the first $n-1$ tasks, the remaining energy level is not less than the energy level required to complete the last task $m_n$, i.e.,

$$ E-\sum_{i=1}^{n-1} a_i \geq m_n $$

We can express $m_n$ as $a_n+(m_n - a_n)$, and then transform the above formula to get:

$$ E-\sum_{i=1}^{n-1} a_i \geq a_n+(m_n - a_n) $$

Rearranging, we get:

$$ E \geq \sum_{i=1}^{n} a_i + (m_n - a_n) $$

Where $\sum_{i=1}^{n} a_i$ is fixed. To minimize the initial energy level $E$, we need to minimize $m_n - a_n$, i.e., maximize $a_n-m_n$.

Therefore, we can sort the tasks in ascending order of $a_i-m_i$. Then we traverse the tasks from front to back. For each task, if the current energy level $cur$ is less than $m_i$, we need to increase the energy level by $m_i - cur$ to make the current energy level exactly equal to $m_i$. Then we complete the task and update $cur = cur - a_i$. Continue traversing until all tasks are completed, and we can get the minimum initial energy level required.

The time complexity is $O(n\times \log n)$, where $n$ is the number of tasks. Ignoring the space overhead of sorting, the space complexity is $O(1)$.

Python3

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

Java

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

C++

class Solution {
public:
    int minimumEffort(vector<vector<int>>& tasks) {
        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;
        for (auto& task : tasks) {
            int a = task[0], m = task[1];
            if (cur < m) {
                ans += m - cur;
                cur = m;
            }
            cur -= a;
        }
        return ans;
    }
};

Go

func minimumEffort(tasks [][]int) (ans int) {
	sort.Slice(tasks, func(i, j int) bool { return tasks[i][0]-tasks[i][1] < tasks[j][0]-tasks[j][1] })
	cur := 0
	for _, task := range tasks {
		a, m := task[0], task[1]
		if cur < m {
			ans += m - cur
			cur = m
		}
		cur -= a
	}
	return
}

TypeScript

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;
}