In this exercise, let's try to solve a classic problem.

Bob is a thief. After months of careful planning, he finally manages to crack the security systems of a high-class apartment.

In front of him are many items, each with a value (v) and weight (w). Bob, of course, wants to maximize the total value he can get; he would gladly take all of the items if he could. However, to his horror, he realizes that the knapsack he carries with him can only hold so much weight (W).

Given a knapsack with a specific carrying capacity (W), help Bob determine the maximum value he can get from the items in the house. Note that Bob can take only one of each item.

All values given will be strictly positive. Items will be represented as a list of pairs, wi and vi, where the first element wi is the weight of the ith item and vi is the value for that item.

For example:

Items: [ { "weight": 5, "value": 10 }, { "weight": 4, "value": 40 }, { "weight": 6, "value": 30 }, { "weight": 4, "value": 50 } ]

Knapsack Limit: 10

For the above, the first item has weight 5 and value 10, the second item has weight 4 and value 40, and so on.

In this example, Bob should take the second and fourth item to maximize his value, which, in this case, is 90. He cannot get more than 90 as his knapsack has a weight limit of 10.


WikipediaThe link opens in a new window or tab
Last updated 15 September 2021
Edit via GitHub The link opens in a new window or tab
Java Exercism

Ready to start Knapsack?

Sign up to Exercism to learn and master Java with 17 concepts, 128 exercises, and real human mentoring, all for free.