1784 Minimum Initial Energy To Finish Tasks

Minimum Initial Energy to Finish Tasks

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

impl Solution {
    pub fn minimum_effort(mut tasks: Vec<Vec<i32>>) -> i32 {
        tasks.sort_by(|a, b| (a[1] - a[0]).cmp(&(b[1] - b[0])));
        let mut ans = 0;
        for task in tasks.iter() {
            ans = std::cmp::max(ans + task[0], task[1]);
        }
        ans
    }
}