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Extended Knapsack Problem

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  • Difficulty Level : Medium
  • Last Updated : 03 Oct, 2022
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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:
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 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 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 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-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
vector<vector<vector<int> > > dp;
 
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()
{
    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;
 
    dp = vector<vector<vector<int> > >(
        n + 1, vector<vector<int> >(
                   max_weight + 1,
                   vector<int>(max_element + 1, 0)));
    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


Python3




# Python3 code for the extended
# Knapsack Approach
 
# To store the dp values
dp=[]
 
def maxProfit(profit, weight, n, max_W,
              max_E):
 
    # for each element given
    for i in range(1,n+1) :
 
        # For each possible
        # weight value
        for j in range(1,max_W+1) :
 
            # For each case where
            # the total elements are
            # less than the constra
            for k in range(1, max_E+1) :
 
                # 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
if __name__ == '__main__':
    n = 5
    profit = [2, 7, 1, 5, 3 ]
    weight = [ 2, 5, 2, 3, 4 ]
    max_weight = 8
    max_element = 2
 
    dp = [[[0 for j in range(max_element + 1)]for i in range(max_weight + 1)] for k in range(n+1)]
    print(maxProfit(profit, weight, n, max_weight,
                      max_element))


C#




// C# program to implement above approach
using System;
using System.Collections;
using System.Collections.Generic;
 
class GFG
{
 
  // To store the dp values
  static int[][][] dp = new int[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 = new int[]{ 2, 7, 1, 5, 3 };
    int[] weight = new int[]{ 2, 5, 2, 3, 4 };
    int max_weight = 8;
    int max_element = 2;
 
    // Initialize dp
    for(int i = 0 ; i < 100 ; i++){
      dp[i] = new int[100][];
      for(int j = 0 ; j < 100 ; j++){
        dp[i][j] = new int[100];
      }
    }
 
    Console.WriteLine(maxProfit(profit, weight, n,
                                max_weight, max_element));
 
  }
}
 
// This code is contributed by subhamgoyal2014.


Javascript




<script>
 
// Javascript code for the extended
// Knapsack Approach
 
// To store the dp values
var dp = Array.from(Array(100), ()=>Array(100));
for(var i =0; i<100; i++)
        for(var j =0; j<100; j++)
            dp[i][j] = new Array(100).fill(0);
 
function maxProfit(profit,weight, n, max_W, max_E)
{
 
    // for each element given
    for (var i = 1; i <= n; i++)
    {
 
        // For each possible
        // weight value
        for (var j = 1; j <= max_W; j++)
        {
 
            // For each case where
            // the total elements are
            // less than the constraint
            for (var 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
var n = 5;
var profit = [2, 7, 1, 5, 3 ];
var weight = [2, 5, 2, 3, 4 ];
var max_weight = 8;
var max_element = 2;
document.write( maxProfit(profit,
              weight, n,
              max_weight,
              max_element)
     + "<br>");
 
 
</script>


Output: 

12

 

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

Approach: 2 The above code takes O(N * W * K) extra space, however, this can be reduced to a 2d matrix. Note that only space would be reduced to O(K * W), and time complexity remains the same, i.e. O(N * W * K).

  1. The idea is to eliminate the array storage for items. If you see, we update a cell based on dp[i][j][k] = max( dp[i – 1][j][k],     dp[i – 1][j – weight[i – 1]][k – 1] + profit[i – 1]);
    instead, we can only use a 2d matrix where the x-axis would represent knapsack capacity (0, 1, 2, 3…W) and y-axis represents k items picked (0, 1, 2, 3, … k).
  2. Note that a cell is updated keeping only two values in mind, the previous value dp[i – 1][j][k], and the value for remaining capacity with k – 1 items. So we only need the previous row to check for [i – 1 ] and [i – 1][j – weight[i – 1]].
  3. A cell dp[j][k] in the matrix represents the maximum profit we can get with capacity j, if at most k, items are allowed, with all the items till i. This way we can still iterate over items but we don’t need to keep storage for them.

Below is the implementation of the above approach:

C++




#include <bits/stdc++.h>
#include <iostream>
 
using namespace std;
 
int maxProfit(int profit[], int weight[], int n,
              int maxCapacity, int maxItems)
{
    vector<vector<int> > dp(
        maxItems + 1, vector<int>(maxCapacity + 1, 0));
 
    for (int i = 0; i < n; i++) {
        for (int k = 1; k < maxItems + 1; k++) {
            for (int j = maxCapacity; j >= 0; j--) {
                if (weight[i] <= j) {
                    dp[k][j] = max(
                        dp[k][j],
                        profit[i]
                            + dp[k - 1][j - weight[i]]);
                }
            }
        }
    }
    return dp[maxItems][maxCapacity];
}
 
// Drivers Code
int main()
{
 
    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;
 
      // Function Call
    cout << maxProfit(profit, weight, n, max_weight,
                      max_element)
         << "\n";
    return 0;
}


Output

12

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


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