2244.Minimum Rounds to Complete All Tasks
===

###### tags: `Medium`,`Array`,`Hash Table`,`Greedy`,`Counting`

[2244. Minimum Rounds to Complete All Tasks](https://leetcode.com/problems/minimum-rounds-to-complete-all-tasks/)

### 題目描述

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.*

### 範例

**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` <= 10^5^
* 1 <= `tasks[i]` <= 10^9^

### 解答
#### Python

class Solution:
    def minimumRounds(self, tasks: List[int]) -> int:
        counter = Counter(tasks)
        ans = 0
        
        for cnt in counter.values():
            if cnt < 2:
                return -1
            else:
                ans += (cnt + 2) // 3

        return ans

寫到這靈感就來了

class Solution:
    def minimumRounds(self, tasks: List[int]) -> int:
        return -1 if 1 in Counter(tasks).values() else sum((cnt + 2) // 3 for cnt in Counter(tasks).values())

> [name=Kobe][time=Jan 4, 2023]

### Reference

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