LeetCode每日一题(2244. Minimum Rounds to Complete All Tasks)

You are given a 0-indexed integer array tasks, where tasks[i] represents the difficulty level of a task. In each round, you can complete either 2 or 3 tasks of the same difficulty level.

Return the minimum rounds required to complete all the tasks, or -1 if it is not possible to complete all the tasks.

Example 1:

Input: tasks = [2,2,3,3,2,4,4,4,4,4]
Output: 4

Explanation: To complete all the tasks, a possible plan is:

• In the first round, you complete 3 tasks of difficulty level 2.
• In the second round, you complete 2 tasks of difficulty level 3.
• In the third round, you complete 3 tasks of difficulty level 4.
• In the fourth round, you complete 2 tasks of difficulty level 4.
  It can be shown that all the tasks cannot be completed in fewer than 4 rounds, so the answer is 4.

Example 2:

Input: tasks = [2,3,3]
Output: -1

Explanation: There is only 1 task of difficulty level 2, but in each round, you can only complete either 2 or 3 tasks of the same difficulty level. Hence, you cannot complete all the tasks, and the answer is -1.

Constraints:

• 1 <= tasks.length <= 105
• 1 <= tasks[i] <= 109

---

对于难度 d, 如果:

• task 数量为 1, 因为我们只能同时处理 2 或者 3 个 task， 所以该 task 我们无法完成， 返回-1
• task 数量为 2, 1 次即可完成
• task 数量为 3, 1 次即可完成
• task 数量>=4, 为了使次数最少， 我们应该优先同时处理 3 个 task，但是如果 num % 3 == 1, 那就意味着剩余的那 1 个 task 我们无法处理， 我们把这种情况罗列一下， 4, 7, 10, 13…, 我们可以发现这些数字都可以拆解成 4 + n * 3 的形式， 4 我们可以用 2 步处理完， 所以处理这些数字的步数就是(x - 4) / 3 + 2

---

use std::collections::HashMap;

impl Solution {
    pub fn minimum_rounds(tasks: Vec<i32>) -> i32 {
        let m = tasks.into_iter().fold(HashMap::new(), |mut m, t| {
            *m.entry(t).or_insert(0) += 1;
            m
        });
        let mut counts: Vec<i32> = m.values().map(|&v| v).collect();
        counts.sort();
        let mut ans = 0;
        for c in counts {
            match c {
                1 => return -1,
                2 => ans += 1,
                3 => ans += 1,
                _ => {
                    ans += (c - 4) / 3 + 2;
                }
            }
        }
        ans
    }
}