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Formatted question description: https://leetcode.ca/all/2244.html

# 2244. Minimum Rounds to Complete All Tasks (Medium)

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}

**Similar Questions**:

## Solution 1.

```
// OJ: https://leetcode.com/problems/minimum-rounds-to-complete-all-tasks/
// Time: O(N)
// Space: O(U) where `U` is the count of distinct numbers in `A`.
class Solution {
public:
int minimumRounds(vector<int>& A) {
unordered_map<int, int> m;
for (int n : A) m[n]++;
int ans = 0;
for (auto &[n, cnt] : m) {
if (cnt % 3 == 0) ans += cnt / 3;
else if (cnt % 3 == 1) {
if (cnt == 1) return -1;
ans += 2 + (cnt - 4) / 3;
} else ans += 1 + (cnt - 1) / 3;
}
return ans;
}
};
```

Or

```
// OJ: https://leetcode.com/problems/minimum-rounds-to-complete-all-tasks/
// Time: O(N)
// Space: O(U) where `U` is the count of distinct numbers in `A`.
class Solution {
public:
int minimumRounds(vector<int>& A) {
unordered_map<int, int> m;
for (int n : A) m[n]++;
int ans = 0;
for (auto &[n, cnt] : m) {
if (cnt == 1) return -1;
ans += (cnt + 2) / 3;
}
return ans;
}
};
```