W

S

```
from bisect import bisect
from itertools import accumulate
def fracKnapsack(vl, wt, W, n):
"""
>>> fracKnapsack([60, 100, 120], [10, 20, 30], 50, 3)
240.0
"""
r = list(sorted(zip(vl, wt), key=lambda x: x[0] / x[1], reverse=True))
vl, wt = [i[0] for i in r], [i[1] for i in r]
acc = list(accumulate(wt))
k = bisect(acc, W)
return (
0
if k == 0
else sum(vl[:k]) + (W - acc[k - 1]) * (vl[k]) / (wt[k])
if k != n
else sum(vl[:k])
)
if __name__ == "__main__":
import doctest
doctest.testmod()
```

Given a set of items, each with weight and a value, determine the number of each item included in a collection so that the total weight is less than or equal to the given limit and the total value is as large as possible.

O(nlog n) Worst Case

```
Lets assume the capacity of knapsack, W = 60
value = [280, 100, 120, 120]
weight = [40, 10, 20, 24]
Ratio(V/W) = 7,10,6,5
Say those items as A,B,C,D
next, the items should be sorted in descending order based on the ratio of value by weight to get maximum profit
First and foremost, B was picked since its weight is smaller than the knapsack's capacity. The next item, A, is chosen since the knapsack's available capacity is more than A's weight. C is now the next item on the list. However, the entire item cannot be chosen because the knapsack's remaining capacity is less than C's weight.
As a result, the C proportion (60ā50)/20)
The knapsack's capacity is now equal to the specified items.
As a result, no more items can be chosen.
10 + 40 + 20*(10/20) = 60 is the total weight of the chosen goods.
100+280+120*(10/20)=380+60=440 is the total profit.
This is the most suitable option.
We won't be able to make more money by combining diverse things.
```