# Solving Leetcode Interviews in Seconds with AI: Minimum Rounds to Complete All Tasks


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "2244" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> 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. 

	# Explanation
	Here's the breakdown of the solution:

*   **Count Task Occurrences:** First, count the occurrences of each difficulty level using a dictionary (or hash map).
*   **Greedy Round Calculation:** For each difficulty level, determine the minimum rounds needed. Prefer completing tasks in groups of 3 whenever possible, then use groups of 2 for the remainder. If a difficulty level has only one task, it's impossible to complete.

*   **Runtime Complexity:** O(n), where n is the number of tasks, due to iterating through the tasks once to count occurrences and then iterating through the distinct difficulty levels.
*   **Storage Complexity:** O(k), where k is the number of distinct difficulty levels. In the worst case, k could be equal to n.

	
	# Code
	```python
	def minimumRounds(tasks: list[int]) -> int:
    """
    Calculates the minimum rounds required to complete all tasks, where each round
    can complete 2 or 3 tasks of the same difficulty level.

    Args:
        tasks: A list of integers representing the difficulty level of each task.

    Returns:
        The minimum rounds required, or -1 if it's not possible.
    """
    task_counts = {}
    for task in tasks:
        task_counts[task] = task_counts.get(task, 0) + 1

    rounds = 0
    for count in task_counts.values():
        if count == 1:
            return -1
        rounds += (count + 2) // 3  # Efficiently calculate rounds

    return rounds
	```