If you just solve subset sum, you're left with 5,5,5 in one bag and 9 in one bag each (for a total of 4 bags), rather than 5,9 in each of 3 bags. Why not? Consider items 5,5,5,9,9,9 with a bag size of 16. For this, you can consider the weight the same as value and solve using a knapsack algorithm, but this won't really work too well for multiple bags. In the case of 1 bag, this is essentially the subset sum problem. There are many types of backpacks: rucksack, knapsack, packsack, pack, etc. The running time will greatly depend on the number of items that can fit into a bag - it will be O(minimumBagsUsed.2 maxItemsPerBag). Key Difference: A backpack is a type of bag that is carried on ones back. The linked resource actually considers multiple types, which can occur multiple times - I derived the above solution from that. The array index above is literally a set - think of this as a map of set to value, a bitmap or a multi-dimensional array where each index is either 1 or 0 to indicate whether we include the item corresponding to that dimensional or not. The Leather Knapsack Backpack, guaranteed for life and handmade in America of natural, vegetable-tanned leather, is a modern, functional bag for work or.
This resource has one option - the basic idea is: D where set2 fits into 1 bag In terms of optimal solutions, there isn't a dynamic programming solution that's as well-known as for the knapsack problem. This would easily take O(n²), or possibly O(n log n) with an efficient implementation. Get a daypack for shorter trips or a heavy duty bag for the long-haul. This is known as the bin packing problem (which is NP-hard).īy simply sorting the decreasing order by their sizes, and then inserting each item into the first bin in the list with sufficient remaining space, we get 11/9 OPT + 6/9 bins (where OPT is the number of bins used in the optimal solution).