Problem Statement

We are going to deal with the well known knapsack problem with an additional constraint in this blog. We are given a list of N items and a knapsack of size W. Every item has a cost c_i associated with it (1 ≤ i ≤ N). We can select some items from the list such sum of the cost of all the selected items does not exceed W. The goal is tell for all w (0 ≤ w ≤ W), if we can select any number of items such that their total cost equals w. This is also known as the 0/1 knapsack problem. The addition constraint we have is .

The bounded knapsack problem

The bounded knapsack problem is like the 0/1 knapsack problem, except in this we are also given a count for each item. In other words, each item has a count s_i associated with it and we can select an item s_i times (1 ≤ i ≤ N).

Solving bounded knapsack problem

The solution is simple. Let dp[i][j] be the minimum count of ith item that has to be used to get a total cost of j while using some number (possibly 0) of first i items. If a total cost of j can not be obtained using first i items, then dp[i][j] =  - 1. The following code is used to calculate the dp[i][j],

if(dp[i-1][j] >= 0)
    dp[i][j] = 0;
else if(dp[i][j - c[i]] >= 0 and dp[i][j - c[i]] < s[i])
    dp[i][j] = dp[i][j - c[i]] + 1;
else
    dp[i][j] = -1;

Here, c[i] is the cost and s[i] is the count for ith item. Also, dp[0][j] =  - 1 for all 1 ≤ j ≤ W and dp[0][0] = 0.