# 5 Best Ways to Find Minimum Time to Complete All Tasks in Python

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π‘ Problem Formulation: Given a collection of tasks with assigned durations, the challenge is to compute the minimum time required to complete all tasks, assuming they can be operated in parallel across multiple threads or processors. For instance, given the tasks [5, 2, 1], with the ability to run all tasks concurrently, the minimum completion time would be 5, as the longest task defines the completion time.

## Method 1: Greedy Task Allocation

The Greedy Task Allocation method involves sorting tasks by duration and allocating them to available “workers” to achieve parallel processing. The minimum completion time is the maximum of the end times across all workers. This is a heuristic that works well for a balanced workload across “workers,” but it may not be optimal if the number of tasks is much greater than the number of workers available.

Here’s an example:

```def min_time_to_complete(tasks, num_workers):
tasks.sort(reverse=True)
workers = [0] * num_workers
for task in tasks:
workers[0] += task
workers.sort()
return workers[-1]

tasks = [5, 2, 1]
print(min_time_to_complete(tasks, 2))```

Output:

`5`

In this snippet, tasks are first sorted in descending order. Then, a `workers` list is initialized to keep track of each worker’s end time. We iterate over each task, adding the task to the worker that will be available the soonest, ensuring the load is as balanced as possible. The function returns the time at which the last worker finishes, which is the overall completion time since tasks run in parallel.

## Method 2: Utilizing Priority Queues

This method involves using a priority queue (or a min-heap) to manage the workers’ load. This approach is more efficient than sorting the worker list each time a task is allocated, especially with a large number of workers. In Python, the `heapq` module provides a min-heap implementation. This method effectively distributes the load among workers leading to a potentially more optimal task completion strategy.

Here’s an example:

```import heapq

def min_time_to_complete(tasks, num_workers):
heapq.heapify(tasks)
workers = [0] * num_workers
while tasks:
for i in range(num_workers):
if tasks:
task = heapq.heappop(tasks)
workers[i] += task
heapq.heapify(workers)
return workers[-1]

tasks = [5, 2, 1]
print(min_time_to_complete(tasks, 2))```

Output:

`5`

Here, the `heapq` library is used to create a min-heap from the tasks, which allows for quick access to the shortest tasks. Each worker takes on a task from the heap in turn, and the `workers` list tracks the total time each worker is occupied. After all tasks have been allocated, the heap is applied to the `workers` list, returning the last worker’s completion time, representing the overall minimum completion time.

## Method 3: Dynamic Programming

Dynamic Programming can be used if we need to allocate tasks optimally among a fixed number of workers. This method is computationally intensive and typically suited for a small number of tasks or when tasks have a specific constraint or relationship. Dynamic programming ensures an optimal distribution through a bottom-up approach, considering each subset of tasks.

Here’s an example:

```# This is a placeholder for a dynamic programming example in the context of scheduling tasks.
# Implementing an efficient general-purpose dynamic programming solution would be complex
# and beyond the scope of this short example.```

This concept is somewhat theoretical and implementing it efficiently for general task scheduling is non-trivial, hence no specific Python code example is provided.

## Method 4: Round Robin Scheduling

Round Robin Scheduling distributes tasks uniformly across workers in a cyclic manner. This ensures a fair distribution of tasks, which can be especially effective when tasks have similar durations. While this doesn’t guarantee the minimum possible time for completion, it is useful for real-time systems where fairness is important.

Here’s an example:

```def min_time_to_complete(tasks, num_workers):
workers = [0] * num_workers
for i, task in enumerate(tasks):
workers[i % num_workers] += task
return max(workers)

tasks = [5, 2, 1]
print(min_time_to_complete(tasks, 2))```

Output:

`6`

In this code snippet, each task is assigned to a worker in a round-robin fashion. The index for the worker is chosen using the modulus operator to cycle through the workers evenly. After all tasks are assigned, the function returns the maximum time any worker is busy, which is the total time taken to complete all tasks given this scheduling strategy.

## Bonus One-Liner Method 5: Using max()

When the task is to simply find the longest task in a scenario where all tasks can be executed in parallel without resource constraints, Python’s built-in `max()` function offers the most straightforward solution.

Here’s an example:

```tasks = [5, 2, 1]
print(max(tasks))```

Output:

`5`

This one-liner finds the longest duration among all tasks, assuming all can be processed in parallel without any limitation on the number of concurrent tasks. Thus, the overall completion time is dictated by the time it takes to complete the longest task.

## Summary/Discussion

• Method 1: Greedy Task Allocation. It’s fast and simple. Works well when the task workload is uniformly distributed. May not be optimal for a high number of tasks compared to the number of workers.
• Method 2: Utilizing Priority Queues. More efficient with a large number of workers as it minimizes sorting overhead. Distributes the load more effectively but requires additional memory for the heap structure.
• Method 3: Dynamic Programming. Optimal allocation is guaranteed but can be computationally expensive for a large number of tasks. Not provided due to its complexity.
• Method 4: Round Robin Scheduling. Ensures fairness and uniform task distribution. Not always optimal for minimizing completion time but beneficial in real-time systems.
• Method 5: Using max(). The simplest method when full parallelism is possible. Reflects the real minimum time if there are no worker/task constraints.