Minimum Rounds to Complete All Tasks

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 <= 10<sup>5</sup>

• 1 <= tasks[i] <= 10<sup>9</sup>

Solution:

Language: Java

class Solution 
{
    public int minimumRounds(int[] tasks) 
    {
        Map<Integer, Integer> freq = new HashMap();

        for (int task : tasks) 
        {
            freq.put(task, freq.getOrDefault(task, 0) + 1);
        }

        int minimumRounds = 0;
        // Iterate over the task's frequencies.
        for (int count : freq.values()) 
        {
            // If the frequency is 1, it's not possible to complete tasks.
            if (count == 1) 
            {
                return - 1;
            }

            if (count % 3 == 0) 
            {
                // Group all the task in triplets.
                minimumRounds += count / 3;
            } 
            else 
            {
                // If count % 3 = 1; (count / 3 - 1) groups of triplets and 2 pairs.
                // If count % 3 = 2; (count / 3) groups of triplets and 1 pair.
                minimumRounds += count / 3 + 1;
            }
        }

        return minimumRounds;
    }
}