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

所有的任務(wù)都要完成，那么我們就去計算每個任務(wù)出現(xiàn)的次數(shù)，如果出現(xiàn)的次數(shù)<2，那么就不能完成，這時候就返回-1；

如果都符合要求，那么就去判斷如何用最少的round去完成，那么我們優(yōu)先用3去判斷余數(shù)，余數(shù)為0，直接全部3， 余數(shù)為1，那么相當(dāng)于最后4個任務(wù)要用2次round去完成，

同理余數(shù)是2的時候，最后2個任務(wù)用1次round完成；

下面就是代碼了，主要就是用hashmap即可；

Runtime:?40 ms, faster than?36.31%?of?Java?online submissions for?Minimum Rounds to Complete All Tasks.

Memory Usage:?56.8 MB, less than?61.18%?of?Java?online submissions for?Minimum Rounds to Complete All Tasks.