Extended Knapsack Problem

Difficulty Level : Hard
Last Updated : 19 Nov, 2020

Given N items, each item having a given weight C_i and a profit value P_i, 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.

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 overlapping solutions.
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 way, the 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 i^th 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-1]][k-1] + P[i])

Below is the implementation of the above approach:

C++

// C++ code for the extended
// Knapsack Approach
#include <bits/stdc++.h>
using namespace std;
 
// To store the dp values
int dp[100][100][100];
 
int maxProfit(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-1]) 
                {
 
                    dp[i][j][k]
                        = max(dp[i - 1][j][k],
                                dp[i - 1][j - 
                          weight[i-1]][k - 1]+ 
                                  profit[i-1]);
                }
                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 << maxProfit(profit,
                  weight, n,
                  max_weight,
                  max_element)
         << "\n";
 
    return 0;
}

Java

// Java code for the extended
// Knapsack Approach
import java.util.*;
import java.lang.*;
import java.io.*;
 
class GFG{
 
// To store the dp values
static int[][][] dp = new int[100][100][100];
 
static int maxProfit(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 - 1]) 
                {
                    dp[i][j][k] = Math.max(dp[i - 1][j][k],
                                           dp[i - 1][j - 
                                       weight[i - 1]][k - 1] +
                                       profit[i - 1]);
                }
                else
                {
                    dp[i][j][k] = dp[i - 1][j][k];
                }
            }
        }
    }
 
    return dp[n][max_W][max_E];
}
   
// Driver code
public static void main(String[] args) 
{
    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;
     
    System.out.println(maxProfit(profit,
                                 weight, n,
                                 max_weight,
                                 max_element));   
}
}
 
// This code is contributed by offbeat

Output:

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

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes

Related Articles

C++

Java