# 5 Best Ways to Find Maximum Time to Finish K Tasks in Python

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π‘ Problem Formulation: Python programmers often encounter the problem of calculating the maximum time required to complete a certain number of tasks, denoted as ‘k’. Suppose you have a list of individual task durations and you want to find out the total time it will take to finish ‘k’ number of these tasks. For instance, if task durations are [5, 2, 3] and k is 2, the maximum time to finish ‘k’ tasks would be 5 + 3 = 8.

## Method 1: Brute Force Approach

This approach involves checking all possible combinations of ‘k’ tasks to determine which combination yields the maximum time. Itβs effective for small datasets but may become impractical as the number of tasks increases due to its O(n^k) time complexity, where n is the number of tasks.

Here’s an example:

```from itertools import combinations

return max(sum(combo) for combo in combinations(tasks, k))

# Example usage
k = 2
```

Output: 8

This Python function `max_time` takes a list of tasks durations and an integer ‘k’, creating all possible combinations of k tasks using the `combinations` method from the itertools module. It then calculates the sum of each combination’s durations and returns the maximum sum.

## Method 2: Sorting and Selecting

By sorting the task durations in descending order and selecting the first ‘k’ elements, we can efficiently determine the maximum time for ‘k’ tasks. This method significantly reduces the time complexity to O(n log n) due to the sorting operation.

Here’s an example:

```def max_time(tasks, k):

# Example usage
k = 2
```

Output: 8

The `max_time` function first sorts the task durations in reverse order to get a descending list, then slices off the first ‘k’ tasks and sums their durations to find the maximum time.

## Method 3: Using a Priority Queue

This method improves on the sorting approach by using a max-heap or priority queue to efficiently manage tasks. With this, the overall complexity can drop to O(k log n), suitable for larger datasets where k is significantly smaller than the number of tasks.

Here’s an example:

```import heapq

max_heap = [-x for x in tasks] # Create a max-heap
heapq.heapify(max_heap)
return -sum(heapq.heappop(max_heap) for _ in range(k))

# Example usage
k = 2
```

Output: 8

The `max_time` function converts the list into a max-heap by negating the task durations, then uses `heapq.heappop()` to retrieve and sum the ‘k’ largest tasks. This is one of the most efficient ways to solve the problem when ‘k’ is much smaller than the total number of tasks.

## Method 4: Using Counter and Heap

This method offers the best of both the sorting and heap approaches for datasets where task durations might frequently repeat. It counts the frequency of each task duration and then uses a max-heap to efficiently find the ‘k’ largest durations. The method also provides increased efficiency for larger datasets.

Here’s an example:

```from heapq import nlargest
from collections import Counter

# Example usage
tasks = [5, 2, 5, 3]
k = 2
```

Output: 10

The `max_time` function counts the task durations using `Counter` and then finds the ‘k’ largest keys in the count dictionary using `nlargest`, where the key is a lambda function that uses both the frequency and the value. This combines frequency and value to efficiently find the maximum time.

## Bonus One-Liner Method 5: Using Lambda and Sorted

This one-liner is for those who love concise Python code. It sorts the task durations and picks the last ‘k’ elements to sum, thereby finding the maximum time. Note, this method is just a condensed version of Method 2.

Here’s an example:

```max_time = lambda tasks, k: sum(sorted(tasks)[-k:])

# Example usage
k = 2
In this snippet, `max_time` is a lambda function that sorts tasks and adds the last ‘k’ elements, providing a quick and elegant solution to the problem. It is ideal for small datasets and simple scenarios.