Minimum Processing Time

Medium

Nutanix

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Topics:

ArraysGreedy Algorithms

You have a certain number of processors, each having 4 cores. The number of tasks to be executed is four times the number of processors. Each task must be assigned to a unique core, and each core can only be used once.

You are given an array processorTime representing the time each processor becomes available and an array tasks representing how long each task takes to complete. Return the minimum time needed to complete all tasks.

Example 1:

Input: processorTime = [8,10], tasks = [2,2,3,1,8,7,4,5]

Output: 16

Explanation:

Assign the tasks at indices 4, 5, 6, 7 to the first processor which becomes available at time = 8, and the tasks at indices 0, 1, 2, 3 to the second processor which becomes available at time = 10.

The time taken by the first processor to finish the execution of all tasks is max(8 + 8, 8 + 7, 8 + 4, 8 + 5) = 16.

The time taken by the second processor to finish the execution of all tasks is max(10 + 2, 10 + 2, 10 + 3, 10 + 1) = 13.

Example 2:

Input: processorTime = [10,20], tasks = [2,3,1,2,5,8,4,3]

Output: 23

Explanation:

Assign the tasks at indices 1, 4, 5, 6 to the first processor and the others to the second processor.

The time taken by the first processor to finish the execution of all tasks is max(10 + 3, 10 + 5, 10 + 8, 10 + 4) = 18.

The time taken by the second processor to finish the execution of all tasks is max(20 + 2, 20 + 1, 20 + 2, 20 + 3) = 23.

Constraints:

1 <= n == processorTime.length <= 25000
1 <= tasks.length <= 10⁵
0 <= processorTime[i] <= 10⁹
1 <= tasks[i] <= 10⁹
tasks.length == 4 * n

Try coding it in LeetCode

Minimum Processing Time

Medium

Nutanix

1 view

Topics:

ArraysGreedy Algorithms

You have a certain number of processors, each having 4 cores. The number of tasks to be executed is four times the number of processors. Each task must be assigned to a unique core, and each core can only be used once.

You are given an array processorTime representing the time each processor becomes available and an array tasks representing how long each task takes to complete. Return the minimum time needed to complete all tasks.

Example 1:

Input: processorTime = [8,10], tasks = [2,2,3,1,8,7,4,5]

Output: 16

Explanation:

Assign the tasks at indices 4, 5, 6, 7 to the first processor which becomes available at time = 8, and the tasks at indices 0, 1, 2, 3 to the second processor which becomes available at time = 10.

The time taken by the first processor to finish the execution of all tasks is max(8 + 8, 8 + 7, 8 + 4, 8 + 5) = 16.

The time taken by the second processor to finish the execution of all tasks is max(10 + 2, 10 + 2, 10 + 3, 10 + 1) = 13.

Example 2:

Input: processorTime = [10,20], tasks = [2,3,1,2,5,8,4,3]

Output: 23

Explanation:

Assign the tasks at indices 1, 4, 5, 6 to the first processor and the others to the second processor.

The time taken by the first processor to finish the execution of all tasks is max(10 + 3, 10 + 5, 10 + 8, 10 + 4) = 18.

The time taken by the second processor to finish the execution of all tasks is max(20 + 2, 20 + 1, 20 + 2, 20 + 3) = 23.

Constraints:

1 <= n == processorTime.length <= 25000
1 <= tasks.length <= 10⁵
0 <= processorTime[i] <= 10⁹
1 <= tasks[i] <= 10⁹
tasks.length == 4 * n

Try coding it in LeetCode

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 are the constraints on the sizes of `processorTime` and `tasks` arrays? Specifically, what is the maximum possible length of each array?
What is the range of values for elements within `processorTime` and `tasks`? Are they guaranteed to be non-negative integers?
If the number of processors is insufficient to handle all tasks (i.e., `len(processorTime) * 4 < len(tasks)`), what should be returned?
Are the arrays `processorTime` and `tasks` guaranteed to be sorted, or do I need to consider sorting them as part of my solution?
Could you provide a small example input and its expected output to clarify the expected behavior?

Brute Force Solution

Approach

The brute force approach aims to find the absolute shortest time by trying every possible assignment of tasks to processors. We will consider every possible combination of assigning tasks to processors and pick the assignment resulting in the smallest completion time. It's like trying every single option until we find the very best one.

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

Start by considering the first processor. Give it the first task, then the first two tasks, then the first three tasks, and so on, until you've tried giving it all the tasks.
For each of those possibilities of what the first processor is doing, look at the second processor. Again, try giving it one of the remaining tasks, then two, then three, and so on.
Repeat this for all the processors. Each time you give some work to a processor, make sure the work hasn't already been assigned to another processor. You can't give the same job to two people.
When you've assigned all the tasks to all the processors, calculate the time it takes for the entire system to finish. This is the longer time it takes any one processor to finish its tasks.
Keep track of the shortest overall time you've found so far.
Repeat all the above steps, trying every single possible combination of assigning tasks to processors, until you've explored every assignment possibility.
Finally, after trying every combination, the shortest overall time you kept track of is your answer.

Code Implementation

def minimum_processing_time_brute_force(processors, tasks):
    minimum_total_time = float('inf')
    
    def calculate_total_time(task_assignments):
        total_time = 0
        for processor_index in range(len(processors)):
            start_time = processors[processor_index]
            task_index = task_assignments[processor_index]
            total_time += start_time + tasks[task_index]
        return total_time
    
    def find_minimum_time(current_assignment, remaining_tasks):
        nonlocal minimum_total_time
        
        if not remaining_tasks:
            # All tasks assigned: Calculate total time and update minimum if needed.
            total_time = calculate_total_time(current_assignment)
            minimum_total_time = min(minimum_total_time, total_time)
            return
        
        processor_index = len(current_assignment)
        
        #Try assigning each remaining task to the current processor
        for task_index in remaining_tasks:
            new_remaining_tasks = remaining_tasks.copy()
            new_remaining_tasks.remove(task_index)

            #Recursively call function to assign other tasks
            find_minimum_time(current_assignment + [task_index], new_remaining_tasks)
    
    # Initialize the process to find min time. 
    find_minimum_time([], list(range(len(tasks))))
    
    return minimum_total_time

Big(O) Analysis

Time Complexity

O((n^p) * n!) – The brute force approach described explores every possible assignment of n tasks to p processors. The number of ways to assign tasks to processors grows exponentially with the number of tasks and processors. The number of ways to partition a set of n elements into p subsets is related to Stirling numbers of the second kind. Additionally, for each such partitioning, there are n! ways to order the tasks. Since we need to iterate through all these combinations and then for each combination calculate the maximum processing time, which takes O(n) time where n is the number of tasks, the overall time complexity is approximately O((n^p) * n!). The (n^p) portion arises from Stirling numbers of the second kind representing the number of ways to partition the tasks among processors, and n! captures the possible permutations of those tasks.

Space Complexity

O(P^T) – The brute force approach, as described, explores all possible assignments of T tasks to P processors. In the worst-case scenario, this exploration is implicitly managed by a recursive call stack representing the current assignment being evaluated. The depth of this recursion can reach T (number of tasks), and at each level, we might need to explore options for which processor to assign a task to, potentially leading to P branches. Therefore, the implicit call stack represents a tree-like structure with a branching factor related to P and a depth related to T, leading to an auxiliary space usage proportional to P raised to the power of T (P^T). This represents the maximum number of active branches (function calls) that need to be stored at any given moment.

Optimal Solution

Approach

This problem is about pairing processors with tasks to minimize the overall time. The key idea is to match the fastest processors with the most time-consuming tasks to reduce the total processing time. We want to distribute the load as efficiently as possible.

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

First, sort both the processors and the tasks in ascending order.
Then, match each processor with the most time-consuming tasks. Each processor handles a set of tasks.
For each processor, find the longest task duration it is assigned.
The minimum processing time is the largest of the maximum task durations among all the processors.

Code Implementation

def minimum_processing_time(processor_times, task_times):
    processor_times.sort()
    task_times.sort(reverse=True)

    max_time = 0
    number_of_processors = len(processor_times)
    
    # Iterate through processors to pair with tasks
    for i in range(number_of_processors):
        #Calculate the processing time for each pair
        current_time = processor_times[i] + task_times[i]
        
        # Find the maximum processing time
        if current_time > max_time:
            max_time = current_time

    return max_time

Big(O) Analysis

Time Complexity

O(n log n) – Sorting the processors array of size 'n' and the tasks array of size 'n' (assuming k processors and k tasks per processor for a total of n = k*k tasks), both take O(n log n) time. Matching processors to tasks then involves iterating through the processors, and for each processor finding the maximum task time in the assigned tasks which takes O(n) in total. Therefore, the dominant operation is sorting, resulting in a time complexity of O(n log n). The other operations happen in linear time and are dominated by the sorting operation.

Space Complexity

O(1) – The described algorithm primarily sorts the input arrays in place. Beyond the input arrays, the algorithm only requires a few constant-sized variables for iteration and comparison during the matching and maximum finding steps. The space used does not scale with the size of the input arrays of processors and tasks. Therefore, the auxiliary space complexity is constant, O(1).

Edge Cases

Case	How to Handle
Empty processorTime or tasks array	Return 0 immediately since there are no processors or tasks to process.
processorTime array size is 1	Calculate the maximum task time and add it to the processor time.
tasks array size is not a multiple of 4	Ensure that all tasks are processed by padding tasks with a sufficiently small value conceptually if the algorithm depends on fixed sizes.
Large processorTime and tasks values that could cause integer overflow	Use long or appropriate data type to store intermediate calculations and return values to prevent overflow.
Negative values in processorTime or tasks arrays	Return an appropriate error code or throw an exception as negative processing or task times are invalid in this problem context.
Zero values in processorTime or tasks arrays	The algorithm should handle zero values correctly, potentially indicating a processor with no overhead or a task that completes instantly.
processorTime values are all the same	The algorithm should correctly determine the minimum time regardless of the distribution of processor times, even when all values are identical.
Tasks values are highly skewed (e.g., one very large task)	Sorting tasks and pairing the largest tasks with the fastest processors ensures efficient processing even with skewed data.