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:4Explanation: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:-1Explanation: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}

**Solution: Math**

Count the frequency of each level. The only case that can not be finished is 1 task at some level. Otherwise we can always finish it by 2, 3 tasks at a time.

if n = 2: 2 => 1 round

if n = 3: 3 => 1 round

if n = 4: 2 + 2 => 2 rounds

if n = 5: 3 + 2 => 2 rounds

…

if n = 3k, n % 3 == 0 : 3 + 3 + … + 3 = k rounds

if n = 3k + 1, n % 3 == 1 : 3*(k – 1) + 2 + 2 = k + 1 rounds

if n = 3k + 2, n % 3 == 2 : 3*k + 2 = k + 1 rounds

We need (n + 2) / 3 rounds.

Time complexity: O(n)

Space complexity: O(n)

## C++

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// Author: Huahua class Solution { public: int minimumRounds(vector<int>& tasks) { unordered_map<int, int> m; for (int t : tasks) ++m[t]; int ans = 0; for (auto [level, count] : m) { if (count == 1) return -1; ans += (count + 2) / 3; } return ans; } }; |