2362. Minimum Rounds To Complete All Tasks¶

Difficulty: Medium

2362. 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: 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

Note: This question is the same as 2870: Minimum Number of Operations to Make Array Empty.

Solution¶

import java.util.HashMap;
import java.util.Map;

class Solution {
    public int minimumRounds(int[] tasks) {
        int n = tasks.length;
        HashMap<Integer, Integer> map = new HashMap<>();
        for (int ele : tasks)
            map.put(ele, map.getOrDefault(ele, 0) + 1);
        int total = 0;
        for (Map.Entry<Integer, Integer> entry : map.entrySet()) {
            int key = entry.getKey();
            int val = entry.getValue();
            if (val == 1)
                return -1;
            if (val % 3 == 0)
                total += val / 3;
            else if (val % 3 == 1)
                total += (val - 4) / 3 + 2;
            else
                total += (val - 2) / 3 + 1;
        }
        return total;
    }
}

Complexity Analysis¶

Time Complexity: O(?)
Space Complexity: O(?)

Approach¶

Detailed explanation of the approach will be added here