# Extended Knapsack Problem

Given **N **items, each item having a given weight **C _{i}** and a profit value

**P**, the task is to maximize the profit by selecting a maximum of

_{i}**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 have 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 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 **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]][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; ` `#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)*

**Auxillary Space:** *O(N * W * K)*

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