Python/knapsack/recursive_approach_knapsack.py
Christian Clauss c909da9b08
pre-commit: Upgrade psf/black for stable style 2023 (#8110)
* pre-commit: Upgrade psf/black for stable style 2023

Updating https://github.com/psf/black ... updating 22.12.0 -> 23.1.0 for their `2023 stable style`.
* https://github.com/psf/black/blob/main/CHANGES.md#2310

> This is the first [psf/black] release of 2023, and following our stability policy, it comes with a number of improvements to our stable style…

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Co-authored-by: pre-commit-ci[bot] <66853113+pre-commit-ci[bot]@users.noreply.github.com>
2023-02-01 18:44:54 +05:30

52 lines
1.4 KiB
Python

# To get an insight into naive recursive way to solve the Knapsack problem
"""
A shopkeeper has bags of wheat that each have different weights and different profits.
eg.
no_of_items 4
profit 5 4 8 6
weight 1 2 4 5
max_weight 5
Constraints:
max_weight > 0
profit[i] >= 0
weight[i] >= 0
Calculate the maximum profit that the shopkeeper can make given maxmum weight that can
be carried.
"""
def knapsack(
weights: list, values: list, number_of_items: int, max_weight: int, index: int
) -> int:
"""
Function description is as follows-
:param weights: Take a list of weights
:param values: Take a list of profits corresponding to the weights
:param number_of_items: number of items available to pick from
:param max_weight: Maximum weight that could be carried
:param index: the element we are looking at
:return: Maximum expected gain
>>> knapsack([1, 2, 4, 5], [5, 4, 8, 6], 4, 5, 0)
13
>>> knapsack([3 ,4 , 5], [10, 9 , 8], 3, 25, 0)
27
"""
if index == number_of_items:
return 0
ans1 = 0
ans2 = 0
ans1 = knapsack(weights, values, number_of_items, max_weight, index + 1)
if weights[index] <= max_weight:
ans2 = values[index] + knapsack(
weights, values, number_of_items, max_weight - weights[index], index + 1
)
return max(ans1, ans2)
if __name__ == "__main__":
import doctest
doctest.testmod()