Minimum Rounds to Complete All Tasks

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ArraysGreedy AlgorithmsDynamic Programming

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.

For example:

tasks = [2,2,3,3,2,4,4,4,4,4]

Difficulty level 2 has 3 tasks.
Difficulty level 3 has 2 tasks.
Difficulty level 4 has 5 tasks.

A valid solution is:

Round 1: complete 3 tasks of difficulty level 2.
Round 2: complete 2 tasks of difficulty level 3.
Round 3: complete 3 tasks of difficulty level 4.
Round 4: complete 2 tasks of difficulty level 4.

Therefore the minimum rounds to complete all tasks is 4.

tasks = [2,3,3]

Difficulty level 2 has 1 task.
Difficulty level 3 has 2 tasks.

It is impossible to complete the tasks since you can only complete 2 or 3 tasks of the same difficulty in each round, and there is only 1 task of difficulty level 2. Therefore the answer is -1.

Minimum Rounds to Complete All Tasks

Medium

Amazon

2 views

Topics:

ArraysGreedy AlgorithmsDynamic Programming

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.

For example:

tasks = [2,2,3,3,2,4,4,4,4,4]

Difficulty level 2 has 3 tasks.
Difficulty level 3 has 2 tasks.
Difficulty level 4 has 5 tasks.

A valid solution is:

Round 1: complete 3 tasks of difficulty level 2.
Round 2: complete 2 tasks of difficulty level 3.
Round 3: complete 3 tasks of difficulty level 4.
Round 4: complete 2 tasks of difficulty level 4.

Therefore the minimum rounds to complete all tasks is 4.

tasks = [2,3,3]

Difficulty level 2 has 1 task.
Difficulty level 3 has 2 tasks.

It is impossible to complete the tasks since you can only complete 2 or 3 tasks of the same difficulty in each round, and there is only 1 task of difficulty level 2. Therefore the answer is -1.

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the range of values for the task difficulties (i.e., the integers in the input array)? Can the task difficulties be zero or negative?
What should I return if it is impossible to complete all tasks? Should I return -1, throw an exception, or something else?
How large can the input array (tasks) be? Is there a practical upper limit I should consider for memory usage?
Are the integers representing the task difficulty guaranteed to be integers, or could they be floating-point numbers?
Are the tasks guaranteed to be non-empty (i.e., will the input array always contain at least one element)?

Brute Force Solution

Approach

The brute force method for this problem involves exploring every possible combination of how to group tasks of the same difficulty. We will try to form groups of size two or three until all tasks are assigned, keeping track of the number of rounds needed for each possible arrangement.

Here's how the algorithm would work step-by-step:

Consider all possible ways to group tasks of the same difficulty level.
For each difficulty level, start by assuming we only use groups of size two. Check if this is a valid solution.
If using only groups of two works, calculate the number of rounds needed. If not, move on.
Next, try using only groups of size three for that difficulty level. Calculate the number of rounds if it works.
If neither of the above work, try all combinations of groups of two and three for that difficulty level.
Calculate the number of rounds for each combination.
Repeat this process for every different difficulty level.
For each complete arrangement across all difficulty levels, sum up the number of rounds needed.
Keep track of the minimum number of rounds among all valid arrangements.
The minimum value found is the answer.

Code Implementation

def minimum_rounds_brute_force(tasks):
    task_counts = {}
    for task in tasks:
        task_counts[task] = task_counts.get(task, 0) + 1

    min_rounds = float('inf')

    def solve(difficulty_levels, current_rounds):
        nonlocal min_rounds

        if not difficulty_levels:
            min_rounds = min(min_rounds, current_rounds)
            return

        difficulty = next(iter(difficulty_levels))
        count = difficulty_levels[difficulty]

        # Try all combinations of 2s and 3s
        for num_threes in range(count // 3 + 1):
            remaining = count - num_threes * 3

            if remaining % 2 == 0:
                num_twos = remaining // 2
                total_rounds = num_threes + num_twos

                remaining_levels = difficulty_levels.copy()
                del remaining_levels[difficulty]
                # Recurse, accumulating rounds.
                solve(remaining_levels, current_rounds + total_rounds)

    # The core of our brute force approach.
    solve(task_counts, 0)

    if min_rounds == float('inf'):
        return -1
    else:
        return min_rounds

Big(O) Analysis

Time Complexity

O(3^(n/2)) – The algorithm iterates through all possible combinations of groups of size two and three for each difficulty level. For a single difficulty level with n tasks, the number of possible combinations can grow exponentially. In the worst case, we might need to explore all combinations which is in the order of 3^(n/2), because the number of 3 sized groups will be at most n/3 and we check all possible combinations of this choice and this will give an approximate upper bound for single difficulty tasks. Since we repeat this for each difficulty level, in the worst case, we would have to combine these for all possible values of n, increasing the constant factor of 3^(n/2). Thus, the total number of operations approximates 3^(n/2), which simplifies to O(3^(n/2)).

Space Complexity

O(1) – The described brute force method primarily uses iterative calculations and comparisons. While the process involves exploring different grouping combinations, it doesn't explicitly state the creation of any auxiliary data structures that scale with the input size N (number of tasks). The algorithm focuses on recalculating and comparing round counts without persistently storing intermediate combinations or large amounts of data, suggesting constant extra space usage.

Optimal Solution

Approach

The most efficient way to solve this task scheduling problem is to first count how many times each task appears. Then, for each task, figure out the fewest number of rounds needed to complete all instances of it, and add those up.

Here's how the algorithm would work step-by-step:

First, go through the list of tasks and count how many times each unique task appears. For example, if 'A' appears 5 times, we know there are 5 instances of task 'A' to complete.
Next, look at the count for each task. If a task only appears once, it's impossible to complete because we need at least two instances of a task to form a valid round. In this case, we can immediately say that the tasks cannot be completed.
If a task appears two or more times, we want to complete its instances in as few rounds as possible. We can efficiently complete them using rounds of two or three tasks at a time.
For each task, calculate the minimum number of rounds needed to complete all its instances. If the number of instances is divisible by 3, then the number of rounds is simply the number of instances divided by 3. If there's a remainder of 1 or 2 when dividing by 3, we need to adjust the number of rounds. If remainder is one, we can convert one group of 3 into 2 groups of 2 (i.e., 3 + 1 becomes 2 + 2). If the remainder is two we need to have one more round to deal with that two remaining task.
Finally, add up the minimum number of rounds needed for each task. This sum represents the overall minimum number of rounds required to complete all tasks, following the rules.

Code Implementation

def minimum_rounds_to_complete_all_tasks(tasks):
    task_counts = {}
    for task in tasks:
        task_counts[task] = task_counts.get(task, 0) + 1

    total_rounds = 0

    for task, count in task_counts.items():
        if count == 1:
            # Impossible if a task appears only once.
            return -1

        # Calculate rounds needed for each task type.
        if count % 3 == 0:
            total_rounds += count // 3
        elif count % 3 == 1:
            #  Convert one group of 3 into 2 groups of 2.
            total_rounds += (count // 3) - 1 + 2

        else:
            # Remainder is 2, so one more round.
            total_rounds += (count // 3) + 1

    return total_rounds

Big(O) Analysis

Time Complexity

O(n) – The solution iterates through the input list of tasks once to count the frequency of each task, where n is the number of tasks. This step takes O(n) time. Subsequently, it iterates through the unique task counts, performing constant-time arithmetic operations (division, modulo) for each. Since the number of unique tasks is at most n, this step also takes O(n) time. Therefore, the overall time complexity is dominated by the initial frequency counting, resulting in O(n) time complexity.

Space Complexity

O(N) – The primary auxiliary space usage comes from the need to count the occurrences of each task. This is achieved using a hash map (or dictionary) to store the task counts. In the worst-case scenario, all tasks are unique, resulting in a hash map storing N key-value pairs, where N is the number of tasks. Therefore, the space complexity is directly proportional to the number of tasks in the input list.

Edge Cases

Case	How to Handle
Empty tasks array	Return 0 if the task array is empty, as there are no tasks to complete.
Tasks array with only one element	Return -1 if there is only one task of a particular difficulty because tasks must be completed in groups of 2 or 3.
Very large tasks array (scalability)	Ensure the solution uses efficient data structures (e.g., hash map) and algorithms (linear time) to avoid exceeding time limits.
Tasks array with all identical values	Correctly calculate the minimum rounds needed to complete all tasks of the same difficulty, accounting for divisions by 2 and 3.
Tasks array where one task has frequency 1, and all others have frequencies of 2 or 3	The algorithm should return -1 if there exists any task with frequency equal to 1.
Tasks array with extreme frequency values (e.g., a task that appears 100000 times).	Ensure that the integer division and round calculation does not cause overflow or precision errors.
No solution exists (e.g., task frequency of 1)	Return -1 when any task has only one occurrence, as it's impossible to group it with 2 or 3 other tasks of same difficulty.
Tasks array with a large number of different difficulty levels.	The solution needs to manage memory efficiently to store the counts of many distinct difficulty levels.