Extended Knapsack Problem

Given N items, each item having a given weight Ci and a profit value Pi, the task is to maximize the profit by selecting a maximum of K items adding up to a maximum weight W.

Examples:

Input: N = 5, P[] = {2, 7, 1, 5, 3}, C[] = {2, 5, 2, 3, 4}, W = 8, K = 2.
Output: 12
Explanation:
Here, the maximum possible profit is when we take 2 items: item2 (P[1] = 7 and C[1] = 5) and item4 (P[3] = 5 and C[3] = 3).
Hence, maximum profit = 7 + 5 = 12

Input: N = 5, P[] = {2, 7, 1, 5, 3}, C[] = {2, 5, 2, 3, 4}, W = 1, K = 2
Output: 0
Explanation: All weights are greater than 1. Hence, no item can be picked.

Approach: The dynamic programming approach is preferred over the general recursion approach. Let us first verify that the conditions of DP are still satisfied.



  1. Overlapping sub-problems: When the recursive solution is tried, 1 item is added first and the solution set is (1), (2), …(n). In the second iteration we have (1, 2) and so on where (1) and (2) are recalculated. Hence there will be have overlapping solutions.
  2. Optimal substructure: Overall, each item has only two choices, either it can be included in the solution or denied. For a particular subset of z elements, the solution for (z+1)th element can either have a solution corresponding to the z elements or the (z+1)th element can be added if it doesn’t exceed the knapsack constraints. Either ways optimal substructure property is satisfied.

Let’s derive the recurrence. Let us consider a 3-dimensional table dp[N][W][K], where N is the number of elements, W is the maximum weight capacity and K is the maximum number of items allowed in the knapsack. Let’s define a state dp[i][j][k] where i denotes that we are considering the ith element, j denotes the current weight filled and k denotes the number of items filled until now.

For every state dp[i][j][k], the profit is either that of the previous state (when the current state is not included) or the profit of the current item added to that of the previous state (when the current item is selected). Hence, the recurrence relation is:

dp[i][j][k] = max( dp[i-1][j][k], dp[i-1][j-W[i]][k-1] + P[i])

Below is the implementation of the above approach:

C++

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// C++ code for the extended
// Knapsack Approach
  
#include <bits/stdc++.h>
using namespace std;
#define N 100
#define W 100
#define K 100
  
int dp[N][W][K];
int solve(int profit[],
          int weight[],
          int n, int max_W,
          int max_E)
{
  
    // for each element given
    for (int i = 1; i <= n; i++) {
  
        // For each possible
        // weight value
        for (int j = 1; j <= max_W; j++) {
  
            // For each case where
            // the total elements are
            // less than the constraint
            for (int k = 1; k <= max_E; k++) {
  
                // To ensure that we dont
                // go out of the array
                if (j >= weight[i]) {
  
                    dp[i][j][k]
                        = max(
                            dp[i - 1][j][k],
                            dp[i - 1]
                              [j - weight[i]]
                              [k - 1]
                                + profit[i]);
                }
                else {
                    dp[i][j][k]
                        = dp[i - 1][j][k];
                }
            }
        }
    }
  
    return dp[n][max_W][max_E];
}
  
// Driver Code
int main()
{
  
    memset(dp, 0, sizeof(dp));
  
    int n = 5;
    int profit[] = { 2, 7, 1, 5, 3 };
    int weight[] = { 2, 5, 2, 3, 4 };
    int max_weight = 8;
    int max_element = 2;
    cout << solve(profit,
                  weight, n,
                  max_weight,
                  max_element)
         << "\n";
  
    return 0;
}

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Output:

12

Time Complexity: O(N * W * K)
Auxiliary Space: O(N * W * K)

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