Understanding the counting knapsack problem

Unlike the 0/1 knapsack problem, many real-world problems ask us not how much value we can pack, but in how many different ways a target can be assembled, allowing the same option to be used an unlimited number of times rather than restricting each choice to a single use. In these situations, the decision is not simply whether to include an option or not, but how many times it should be chosen while still meeting the overall target exactly. Each option has an associated cost, and the objective is to determine the number of distinct combinations that meet the required target without exceeding it.

A fundamental problem in this space is the counting knapsack problem, where "counting" means that we count the number of distinct combinations of items, each of which can be included an unlimited number of times.

The counting knapsack problem extends beyond simple one-time selections and proves essential in scenarios like currency denomination (counting the number of ways to make change for an amount with unlimited coins of each type), rod cutting problems (counting the ways a rod or material can be cut where the same length can be cut repeatedly), production planning (counting the ways items can be manufactured with unlimited raw materials), and resource allocation where the same type of resource can be deployed multiple times.

In this lesson, we will learn about the counting knapsack problem and how it can be solved efficiently using a dynamic programming solution.

The counting knapsack problem

Consider we are given n types of items, where the item at index i has a weight weights[i]. We also have a knapsack with a target weight capacity capacity. Unlike the 0/1 variant, each item can be included zero, one, or multiple times in the knapsack.