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# Minimum Rounds to Complete All Tasks LeetCode Solution

## Problem – Minimum Rounds to Complete All Tasks LeetCode Solution

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`

### Minimum Rounds to Complete All Tasks LeetCode Solution in Java

``````    public int minimumRounds(int[] A) {
HashMap<Integer, Integer> count = new HashMap<>();
for (int a : A)
count.put(a, count.getOrDefault(a, 0) + 1);
int res = 0;
for (int freq : count.values()) {
if (freq == 1) return -1;
res += (freq + 2) / 3;
}
return res;
}
``````

### Minimum Rounds to Complete All Tasks LeetCode Solution in C++

``````    int minimumRounds(vector<int>& A) {
unordered_map<int, int> count;
int res = 0, freq1;
for (int a: A)
++count[a];
for (auto& it: count) {
if (it.second == 1) return -1;
res += (it.second + 2) / 3;
}
return res;
}
``````

### Minimum Rounds to Complete All Tasks LeetCode Solution in Python

``````    def minimumRounds(self, tasks):
return -1 if 1 in freq else sum((a + 2) // 3 for a in freq)
``````
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