# 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 = 7 and C = 5) and item4 (P = 5 and C = 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]][k-1] + P[i])

Note: In code we have used 1 based indexing so, we are doing i – 1 for weight and profit array.

Below is the implementation of the above approach:

## C++

 `// C++ code for the extended` `// Knapsack Approach` `#include ` `using` `namespace` `std;`   `// To store the dp values` `vector > > 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 > >(` `        ``n + 1, vector >(` `                   ``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``[][];`   `  ``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``[];` `      ``for``(``int` `j = 0 ; j < 100 ; j++){` `        ``dp[i][j] = ``new` `int``;` `      ``}` `    ``}`   `    ``Console.WriteLine(maxProfit(profit, weight, n, ` `                                ``max_weight, max_element)); `   `  ``}` `}`   `// This code is contributed by subhamgoyal2014.`

## Javascript

 ``

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 ]][k – 1] + profit[i]);
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 ]].
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 ` `#include `   `using` `namespace` `std;`   `int` `maxProfit(``int` `profit[], ``int` `weight[], ``int` `n,` `              ``int` `maxCapacity, ``int` `maxItems)` `{` `    ``vector > 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;` `}`

## Java

 `public` `class` `Main {`   `    ``static` `int` `maxProfit(``int` `profit[], ``int` `weight[], ``int` `n,` `                         ``int` `maxCapacity, ``int` `maxItems)` `    ``{` `        ``int``[][] dp = ``new` `int``[maxItems + ``1``][maxCapacity + ``1``];`   `        ``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] = Math.max(` `                            ``dp[k][j],` `                            ``profit[i]` `                                ``+ dp[k - ``1``][j - weight[i]]);` `                    ``}` `                ``}` `            ``}` `        ``}` `        ``return` `dp[maxItems][maxCapacity];` `    ``}`   `    ``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 devendra1`

## Python3

 `def` `maxProfit(profit, weight, n, maxCapacity, maxItems):` `    ``dp ``=` `[[``0` `for` `j ``in` `range``(maxCapacity ``+` `1``)] ``for` `i ``in` `range``(maxItems ``+` `1``)]`   `    ``for` `i ``in` `range``(n):` `        ``for` `k ``in` `range``(``1``, maxItems ``+` `1``):` `            ``for` `j ``in` `range``(maxCapacity, ``-``1``, ``-``1``):` `                ``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` `if` `__name__ ``=``=` `"__main__"``:` `    ``n ``=` `5` `    ``profit ``=` `[``2``, ``7``, ``1``, ``5``, ``3``]` `    ``weight ``=` `[``2``, ``5``, ``2``, ``3``, ``4``]` `    ``max_weight ``=` `8` `    ``max_element ``=` `2`   `    ``# Function Call` `    ``print``(maxProfit(profit, weight, n, max_weight, max_element))`

## C#

 `using` `System;`   `class` `GFG {`   `  ``// This function calculates the maximum profit` `  ``static` `int` `MaxProfit(``int``[] profit, ``int``[] weight, ``int` `n,` `                       ``int` `maxCapacity, ``int` `maxItems)` `  ``{` `    ``// 2D Array to store DP` `    ``int``[, ] dp = ``new` `int``[maxItems + 1, maxCapacity + 1];`   `    ``// Building the DP array` `    ``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] = Math.Max(` `              ``dp[k, j],` `              ``profit[i]` `              ``+ dp[k - 1, j - weight[i]]);` `          ``}` `        ``}` `      ``}` `    ``}` `    ``return` `dp[maxItems, maxCapacity];` `  ``}`   `  ``// 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;`   `    ``// Function call` `    ``Console.WriteLine(MaxProfit(` `      ``profit, weight, n, max_weight, max_element));` `  ``}` `} `   `// this code is contributed by phasing17`

## Javascript

 `function` `maxProfit(profit, weight, n, maxCapacity, maxItems) {` `  ``// initialize a 2D array with zeros for dynamic programming` `  ``const dp = Array.from(Array(maxItems + 1), () => ``new` `Array(maxCapacity + 1).fill(0));`   `  ``for` `(let i = 0; i < n; i++) {` `    ``for` `(let k = 1; k < maxItems + 1; k++) {` `      ``for` `(let j = maxCapacity; j >= 0; j--) {` `        ``// check if the weight of the item is less than or equal ` `        ``// to the remaining capacity` `        ``if` `(weight[i] <= j) {` `          ``// take the maximum between not including the ` `          ``// current item and including the current item` `          ``dp[k][j] = Math.max(` `            ``dp[k][j], ``// not including the current item` `            ``profit[i] + dp[k - 1][j - weight[i]] ``// including the current item` `          ``);` `        ``}` `      ``}` `    ``}` `  ``}`   `  ``return` `dp[maxItems][maxCapacity]; ``// return the maximum profit achievable` `}`   `// example usage` `const n = 5;` `const profit = [2, 7, 1, 5, 3];` `const weight = [2, 5, 2, 3, 4];` `const max_weight = 8;` `const max_element = 2;`   `console.log(maxProfit(profit, weight, n, max_weight, max_element)); `

Output

```12

```

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

Efficient approach : Space optimization

In previous approach the current value dp[i][j] is only depend upon the current and previous row values of DP. So to optimize the space complexity we use a single 1D array to store the computations.

Implementation steps:

• Create a 1D vector dp of size maxCapacity+1.
• Set a base case by initializing the values of DP .
• Now iterate over subproblems by the help of nested loop and get the current value from previous computations.
• At last return and print the final answer stored in dp[maxCapacity].

Implementation:

## C++

 `#include ` `#include `   `using` `namespace` `std;`   `int` `maxProfit(``int` `profit[], ``int` `weight[], ``int` `n, ``int` `maxCapacity, ``int` `maxItems)` `{` `    ``vector<``int``> dp(maxCapacity + 1, 0);`   `    ``for` `(``int` `i = 0; i < n; i++) {` `        ``for` `(``int` `j = maxCapacity; j >= weight[i]; j--) {` `            ``dp[j] = max(dp[j], profit[i] + dp[j - weight[i]]);` `        ``}` `    ``}` `    ``return` `dp[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;` `}`

## Java

 `import` `java.util.Arrays;`   `class` `GFG {` `    ``public` `static` `int` `maxProfit(``int``[] profit, ``int``[] weight, ``int` `n, ``int` `maxCapacity, ``int` `maxItems) {` `        ``int``[] dp = ``new` `int``[maxCapacity + ``1``];` `        ``Arrays.fill(dp, ``0``);`   `        ``for` `(``int` `i = ``0``; i < n; i++) {` `            ``for` `(``int` `j = maxCapacity; j >= weight[i]; j--) {` `                ``dp[j] = Math.max(dp[j], profit[i] + dp[j - weight[i]]);` `            ``}` `        ``}` `        ``return` `dp[maxCapacity];` `    ``}`   `    ``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``;`   `        ``// Function Call` `        ``System.out.println(maxProfit(profit, weight, n, max_weight, max_element));` `    ``}` `}`

## Python3

 `def` `maxProfit(profit, weight, n, maxCapacity, maxItems):` `    ``dp ``=` `[``0``]``*``(maxCapacity ``+` `1``)` `    ``for` `i ``in` `range``(n):` `        ``for` `j ``in` `range``(maxCapacity, weight[i] ``-` `1``, ``-``1``):` `            ``dp[j] ``=` `max``(dp[j], profit[i] ``+` `dp[j ``-` `weight[i]])` `    ``return` `dp[maxCapacity]`   `# Drivers Code` `if` `__name__ ``=``=` `'__main__'``:` `    ``n ``=` `5` `    ``profit ``=` `[``2``, ``7``, ``1``, ``5``, ``3``]` `    ``weight ``=` `[``2``, ``5``, ``2``, ``3``, ``4``]` `    ``max_weight ``=` `8` `    ``max_element ``=` `2`   `    ``# Function Call` `    ``print``(maxProfit(profit, weight, n, max_weight, max_element))`

## C#

 `using` `System;`   `public` `class` `GFG` `{` `    ``public` `static` `int` `MaxProfit(``int``[] profit, ``int``[] weight, ``int` `n, ``int` `maxCapacity, ``int` `maxItems)` `    ``{` `        ``int``[] dp = ``new` `int``[maxCapacity + 1];`   `        ``for` `(``int` `i = 0; i < n; i++)` `        ``{` `            ``for` `(``int` `j = maxCapacity; j >= weight[i]; j--)` `            ``{` `                ``dp[j] = Math.Max(dp[j], profit[i] + dp[j - weight[i]]);` `            ``}` `        ``}` `        ``return` `dp[maxCapacity];` `    ``}`   `    ``public` `static` `void` `Main(``string``[] args)` `    ``{` `        ``int` `n = 5;` `        ``int``[] profit = { 2, 7, 1, 5, 3 };` `        ``int``[] weight = { 2, 5, 2, 3, 4 };` `        ``int` `maxWeight = 8;` `        ``int` `maxElement = 2;`   `        ``// Function Call` `        ``Console.WriteLine(MaxProfit(profit, weight, n, maxWeight, maxElement));` `    ``}` `}`

## Javascript

 `// Function to calculate the maximum profit for a given knapsack problem` `function` `maxProfit(profit, weight, n, maxCapacity, maxItems) {` `    ``// Create a dp array to store the maximum profit for each capacity (from 0 to maxCapacity)` `    ``const dp = ``new` `Array(maxCapacity + 1).fill(0);`   `    ``// Loop through each item and calculate the maximum profit for each capacity ` `    ``// using bottom-up approach` `    ``for` `(let i = 0; i < n; i++) {` `        ``for` `(let j = maxCapacity; j >= weight[i]; j--) {` `            ``// The maximum profit at capacity 'j' is either the ` `            ``// current profit plus the profit for the remaining capacity (j - weight[i]),` `            ``// or the previous maximum profit at capacity 'j'` `            ``dp[j] = Math.max(dp[j], profit[i] + dp[j - weight[i]]);` `        ``}` `    ``}`   `    ``// The final value in dp array at maxCapacity represents the maximum ` `    ``// profit that can be obtained` `    ``// using the given knapsack capacity and number of items constraint` `    ``return` `dp[maxCapacity];` `}`   `// Drivers Code`   `    ``const n = 5;` `    ``const profit = [2, 7, 1, 5, 3];` `    ``const weight = [2, 5, 2, 3, 4];` `    ``const maxWeight = 8;` `    ``const maxElement = 2;`   `    ``// Function Call` `    ``console.log(maxProfit(profit, weight, n, maxWeight, maxElement));`

Output

`12`

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

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