2244. Minimum Rounds to Complete All Tasks

Problem

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 <= 10^5

• 1 <= tasks[i] <= 10^9

Solution (Java)

class Solution {
    public int minimumRounds(int[] tasks) {
        Arrays.sort(tasks);
        int rounds = 0;
        int c = 1;
        for (int i = 0; i < tasks.length - 1; i++) {
            if (tasks[i] == tasks[i + 1]) {
                c++;
            } else {
                if (c == 1) {
                    return -1;
                }
                if (c % 3 == 0) {
                    rounds += c / 3;
                }
                if (c % 3 >= 1) {
                    rounds += c / 3 + 1;
                }
                c = 1;
            }
        }
        if (c == 1) {
            return -1;
        }
        if (c % 3 == 0) {
            rounds += c / 3;
        }
        if (c % 3 >= 1) {
            rounds += c / 3 + 1;
        }
        return rounds;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).