# Knapsack with large Weights

Given a knapsack with capacity C and two arrays w[] and val[] representing the weights and values of N distinct items, the task is to find the maximum value you can put into the knapsack. Items cannot be broken and an item with weight X takes X capacity of the knapsack.

Examples:

Input: w[] = {3, 4, 5}, val[] = {30, 50, 60}, C = 8
Output: 90
We take objects ‘1’ and ‘3’.
The total value we get is (30 + 60) = 90.
Total weight is 8, thus it fits in the given capacity

Input: w[] = {10000}, val[] = {10}, C = 100000
Output: 10

Approach: The traditional famous 0-1 knapsack problem can be solved in O(N*C) time but if the capacity of the knapsack is huge then a 2D N*C array can’t make be made. Luckily, it can be solved by redefining the states of the dp.
Let’s have a look at the states of the DP first.
dp[V][i] represents the minimum weight subset of the subarray arr[i…N-1] required to get a value of at least V. The recurrence relation will be:

dp[V][i] = min(dp[V][i+1], w[i] + dp[V – val[i]][i + 1])

So, for each V from 0 to the maximum value of V possible, try to find if the given V can be represented with the given array. The largest such V that can be represented becomes the required answer.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach` `#include ` `using` `namespace` `std;`   `#define V_SUM_MAX 1000` `#define N_MAX 100` `#define W_MAX 10000000`   `// To store the states of DP` `int` `dp[V_SUM_MAX + 1][N_MAX];` `bool` `v[V_SUM_MAX + 1][N_MAX];`   `// Function to solve the recurrence relation` `int` `solveDp(``int` `r, ``int` `i, vector<``int``>& w, vector<``int``>& val, ``int` `n)` `{` `    ``// Base cases` `    ``if` `(r <= 0)` `        ``return` `0;` `    ``if` `(i == n)` `        ``return` `W_MAX;` `    ``if` `(v[r][i])` `        ``return` `dp[r][i];`   `    ``// Marking state as solved` `    ``v[r][i] = 1;`   `    ``// Recurrence relation` `    ``dp[r][i]` `        ``= min(solveDp(r, i + 1, w, val, n),` `              ``w[i] + solveDp(r - val[i],` `                             ``i + 1, w, val, n));` `    ``return` `dp[r][i];` `}`   `// Function to return the maximum weight` `int` `maxWeight(vector<``int``>& w, vector<``int``>& val, ``int` `n, ``int` `c)` `{`   `    ``// Iterating through all possible values` `    ``// to find the largest value that can` `    ``// be represented by the given weights` `    ``for` `(``int` `i = V_SUM_MAX; i >= 0; i--) {` `        ``if` `(solveDp(i, 0, w, val, n) <= c) {` `            ``return` `i;` `        ``}` `    ``}` `    ``return` `0;` `}`   `// Driver code` `int` `main()` `{` `    ``vector<``int``> w = { 3, 4, 5 };` `    ``vector<``int``> val = { 30, 50, 60 };` `    ``int` `n = (``int``)w.size();` `    ``int` `C = 8;`   `    ``cout << maxWeight(w, val, n, C);`   `    ``return` `0;` `}`

## Java

 `// Java implementation of the approach` `class` `GFG ` `{` `    ``static` `final` `int` `V_SUM_MAX = ``1000``;` `    ``static` `final` `int` `N_MAX = ``100``;` `    ``static` `final` `int` `W_MAX = ``10000000``;` `    `  `    ``// To store the states of DP ` `    ``static` `int` `dp[][] = ``new` `int``[V_SUM_MAX + ``1``][N_MAX]; ` `    ``static` `boolean` `v[][] = ``new` `boolean` `[V_SUM_MAX + ``1``][N_MAX]; ` `    `  `    ``// Function to solve the recurrence relation ` `    ``static` `int` `solveDp(``int` `r, ``int` `i, ``int` `w[],       ` `                          ``int` `val[], ``int` `n) ` `    ``{ ` `        ``// Base cases ` `        ``if` `(r <= ``0``) ` `            ``return` `0``; ` `            `  `        ``if` `(i == n) ` `            ``return` `W_MAX; ` `            `  `        ``if` `(v[r][i]) ` `            ``return` `dp[r][i]; ` `    `  `        ``// Marking state as solved ` `        ``v[r][i] = ``true``; ` `    `  `        ``// Recurrence relation ` `        ``dp[r][i] = Math.min(solveDp(r, i + ``1``, w, val, n), ` `                     ``w[i] + solveDp(r - val[i], ` `                                    ``i + ``1``, w, val, n)); ` `        `  `        ``return` `dp[r][i]; ` `    ``} ` `    `  `    ``// Function to return the maximum weight ` `    ``static` `int` `maxWeight(``int` `w[], ``int` `val[], ` `                         ``int` `n, ``int` `c) ` `    ``{ ` `    `  `        ``// Iterating through all possible values ` `        ``// to find the largest value that can ` `        ``// be represented by the given weights ` `        ``for` `(``int` `i = V_SUM_MAX; i >= ``0``; i--)` `        ``{ ` `            ``if` `(solveDp(i, ``0``, w, val, n) <= c) ` `            ``{ ` `                ``return` `i; ` `            ``} ` `        ``} ` `        ``return` `0``; ` `    ``} ` `    `  `    ``// Driver code ` `    ``public` `static` `void` `main (String[] args)` `    ``{ ` `        ``int` `w[] = { ``3``, ``4``, ``5` `}; ` `        ``int` `val[] = { ``30``, ``50``, ``60` `}; ` `        ``int` `n = w.length; ` `        ``int` `C = ``8``; ` `    `  `        ``System.out.println(maxWeight(w, val, n, C)); ` `    ``} ` `}`   `// This code is contributed by AnkitRai01`

## Python3

 `# Python3 implementation of the approach` `V_SUM_MAX ``=` `1000` `N_MAX ``=` `100` `W_MAX ``=` `10000000`   `# To store the states of DP` `dp ``=` `[[ ``0` `for` `i ``in` `range``(N_MAX)]` `          ``for` `i ``in` `range``(V_SUM_MAX ``+` `1``)]` `v ``=` `[[ ``0` `for` `i ``in` `range``(N_MAX)]` `         ``for` `i ``in` `range``(V_SUM_MAX ``+` `1``)]`   `# Function to solve the recurrence relation` `def` `solveDp(r, i, w, val, n):` `    `  `    ``# Base cases` `    ``if` `(r <``=` `0``):` `        ``return` `0` `    ``if` `(i ``=``=` `n):` `        ``return` `W_MAX` `    ``if` `(v[r][i]):` `        ``return` `dp[r][i]`   `    ``# Marking state as solved` `    ``v[r][i] ``=` `1`   `    ``# Recurrence relation` `    ``dp[r][i] ``=` `min``(solveDp(r, i ``+` `1``, w, val, n), ` `            ``w[i] ``+` `solveDp(r ``-` `val[i], i ``+` `1``,` `                            ``w, val, n))` `    ``return` `dp[r][i]`   `# Function to return the maximum weight` `def` `maxWeight( w, val, n, c):`   `    ``# Iterating through all possible values` `    ``# to find the largest value that can` `    ``# be represented by the given weights` `    ``for` `i ``in` `range``(V_SUM_MAX, ``-``1``, ``-``1``):` `        ``if` `(solveDp(i, ``0``, w, val, n) <``=` `c):` `            ``return` `i`   `    ``return` `0`   `# Driver code` `if` `__name__ ``=``=` `'__main__'``:` `    ``w ``=` `[``3``, ``4``, ``5``]` `    ``val ``=` `[``30``, ``50``, ``60``]` `    ``n ``=` `len``(w)` `    ``C ``=` `8`   `    ``print``(maxWeight(w, val, n, C))`   `# This code is contributed by Mohit Kumar`

## C#

 `// C# implementation of the approach` `using` `System;`   `class` `GFG ` `{` `    ``static` `readonly` `int` `V_SUM_MAX = 1000;` `    ``static` `readonly` `int` `N_MAX = 100;` `    ``static` `readonly` `int` `W_MAX = 10000000;` `    `  `    ``// To store the states of DP ` `    ``static` `int` `[,]dp = ``new` `int``[V_SUM_MAX + 1, N_MAX]; ` `    ``static` `bool` `[,]v = ``new` `bool` `[V_SUM_MAX + 1, N_MAX]; ` `    `  `    ``// Function to solve the recurrence relation ` `    ``static` `int` `solveDp(``int` `r, ``int` `i, ``int` `[]w,     ` `                       ``int` `[]val, ``int` `n) ` `    ``{ ` `        ``// Base cases ` `        ``if` `(r <= 0) ` `            ``return` `0; ` `            `  `        ``if` `(i == n) ` `            ``return` `W_MAX; ` `            `  `        ``if` `(v[r, i]) ` `            ``return` `dp[r, i]; ` `    `  `        ``// Marking state as solved ` `        ``v[r, i] = ``true``; ` `    `  `        ``// Recurrence relation ` `        ``dp[r, i] = Math.Min(solveDp(r, i + 1, w, val, n), ` `                     ``w[i] + solveDp(r - val[i], ` `                                    ``i + 1, w, val, n)); ` `        `  `        ``return` `dp[r, i]; ` `    ``} ` `    `  `    ``// Function to return the maximum weight ` `    ``static` `int` `maxWeight(``int` `[]w, ``int` `[]val, ` `                         ``int` `n, ``int` `c) ` `    ``{ ` `    `  `        ``// Iterating through all possible values ` `        ``// to find the largest value that can ` `        ``// be represented by the given weights ` `        ``for` `(``int` `i = V_SUM_MAX; i >= 0; i--)` `        ``{ ` `            ``if` `(solveDp(i, 0, w, val, n) <= c) ` `            ``{ ` `                ``return` `i; ` `            ``} ` `        ``} ` `        ``return` `0; ` `    ``} ` `    `  `    ``// Driver code ` `    ``public` `static` `void` `Main(String[] args)` `    ``{ ` `        ``int` `[]w = { 3, 4, 5 }; ` `        ``int` `[]val = { 30, 50, 60 }; ` `        ``int` `n = w.Length; ` `        ``int` `C = 8; ` `    `  `        ``Console.WriteLine(maxWeight(w, val, n, C)); ` `    ``} ` `}`   `// This code is contributed by 29AjayKumar`

## Javascript

 ``

Output:

`90`

Time Complexity: O(V_sum * N) where V_sum is the sum of all the values in the array val[].
Auxiliary Space : O(V_sum * N) where V_sum is the sum of all the values in the array val[].

Efficient approach : Using DP Tabulation method ( Iterative approach )

The approach to solve this problem is same but DP tabulation(bottom-up) method is better then Dp + memoization(top-down) because memoization method needs extra stack space of recursion calls.

Steps to solve this problem :

• Create a table DP to store the solution of the subproblems.
• Initialize the table with base cases
• Now Iterate over subproblems to get the value of current problem form previous computation of subproblems stored in DP
• At last get the maximum value from DP and return the final solution.

Implementation :

## C++

 `// C++ program for above approach`   `#include ` `using` `namespace` `std;`   `#define V_SUM_MAX 1000` `#define N_MAX 100` `#define W_MAX 10000000`   `// Function to return the maximum weight` `int` `maxWeight(vector<``int``>& w, vector<``int``>& val, ``int` `n,` `              ``int` `c)` `{` `    ``// Initialize dp array` `    ``int` `dp[V_SUM_MAX + 1][N_MAX + 1];` `    ``for` `(``int` `i = 0; i <= V_SUM_MAX; i++)` `        ``for` `(``int` `j = 0; j <= n; j++)` `            ``dp[i][j] = W_MAX;`   `    ``// Base case initialization` `    ``for` `(``int` `i = 0; i <= n; i++)` `        ``dp[0][i] = 0;`   `    ``// iterate over subproblems ans get` `    ``// the current value from previous computation` `    ``for` `(``int` `i = 1; i <= V_SUM_MAX; i++)` `        ``for` `(``int` `j = 1; j <= n; j++)` `            ``dp[i][j] = min(` `                ``dp[i][j - 1],` `                ``(i >= val[j - 1])` `                    ``? w[j - 1] + dp[i - val[j - 1]][j - 1]` `                    ``: W_MAX);`   `    ``// Finding maximum value` `    ``for` `(``int` `i = V_SUM_MAX; i >= 0; i--)` `        ``if` `(dp[i][n] <= c)` `            ``return` `i;`   `    ``return` `0;` `}`   `// Driver code` `int` `main()` `{` `    ``vector<``int``> w = { 3, 4, 5 };` `    ``vector<``int``> val = { 30, 50, 60 };` `    ``int` `n = (``int``)w.size();` `    ``int` `C = 8;`   `    ``cout << maxWeight(w, val, n, C);`   `    ``return` `0;` `}`

## Python3

 `import` `sys`   `V_SUM_MAX ``=` `1000` `N_MAX ``=` `100` `W_MAX ``=` `10000000`   `# Function to return the maximum weight`     `def` `maxWeight(w, val, n, c):` `    ``# Initialize dp array` `    ``dp ``=` `[[W_MAX ``for` `j ``in` `range``(n``+``1``)] ``for` `i ``in` `range``(V_SUM_MAX``+``1``)]`   `    ``# Base case initialization` `    ``for` `i ``in` `range``(n``+``1``):` `        ``dp[``0``][i] ``=` `0`   `    ``# iterate over subproblems ans get` `    ``# the current value from previous computation` `    ``for` `i ``in` `range``(``1``, V_SUM_MAX``+``1``):` `        ``for` `j ``in` `range``(``1``, n``+``1``):` `            ``dp[i][j] ``=` `min``(dp[i][j``-``1``], w[j``-``1``]``+``dp[i``-``val[j``-``1``]]` `                           ``[j``-``1``] ``if` `i >``=` `val[j``-``1``] ``else` `W_MAX)`   `    ``# Finding maximum value` `    ``for` `i ``in` `range``(V_SUM_MAX, ``-``1``, ``-``1``):` `        ``if` `dp[i][n] <``=` `c:` `            ``return` `i`   `    ``return` `0`     `# Driver code` `if` `__name__ ``=``=` `"__main__"``:` `    ``w ``=` `[``3``, ``4``, ``5``]` `    ``val ``=` `[``30``, ``50``, ``60``]` `    ``n ``=` `len``(w)` `    ``C ``=` `8`   `    ``print``(maxWeight(w, val, n, C))`

## Java

 `import` `java.util.*;`   `class` `Main {` `    ``public` `static` `final` `int` `V_SUM_MAX = ``1000``;` `    ``public` `static` `final` `int` `N_MAX = ``100``;` `    ``public` `static` `final` `int` `W_MAX = ``10000000``;`   `    ``// Function to return the maximum weight` `    ``public` `static` `int` `maxWeight(List w,` `                                ``List val, ``int` `n,` `                                ``int` `c)` `    ``{` `        ``// Initialize dp array` `        ``int``[][] dp = ``new` `int``[V_SUM_MAX + ``1``][N_MAX + ``1``];` `        ``for` `(``int` `i = ``0``; i <= V_SUM_MAX; i++) {` `            ``for` `(``int` `j = ``0``; j <= n; j++) {` `                ``dp[i][j] = W_MAX;` `            ``}` `        ``}`   `        ``// Base case initialization` `        ``for` `(``int` `i = ``0``; i <= n; i++) {` `            ``dp[``0``][i] = ``0``;` `        ``}`   `        ``// iterate over subproblems ans get` `        ``// the current value from previous computation` `        ``for` `(``int` `i = ``1``; i <= V_SUM_MAX; i++) {` `            ``for` `(``int` `j = ``1``; j <= n; j++) {` `                ``dp[i][j] = Math.min(` `                    ``dp[i][j - ``1``],` `                    ``(i >= val.get(j - ``1``))` `                        ``? w.get(j - ``1``)` `                              ``+ dp[i - val.get(j - ``1``)]` `                                  ``[j - ``1``]` `                        ``: W_MAX);` `            ``}` `        ``}`   `        ``// Finding maximum value` `        ``for` `(``int` `i = V_SUM_MAX; i >= ``0``; i--) {` `            ``if` `(dp[i][n] <= c) {` `                ``return` `i;` `            ``}` `        ``}`   `        ``return` `0``;` `    ``}`   `    ``// Driver code` `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``List w` `            ``= ``new` `ArrayList<>(Arrays.asList(``3``, ``4``, ``5``));` `        ``List val` `            ``= ``new` `ArrayList<>(Arrays.asList(``30``, ``50``, ``60``));` `        ``int` `n = w.size();` `        ``int` `C = ``8``;`   `        ``System.out.println(maxWeight(w, val, n, C));` `    ``}` `}`

## C#

 `using` `System;`   `public` `class` `KnapsackMaxWeight` `{` `    ``private` `const` `int` `V_SUM_MAX = 1000;` `    ``private` `const` `int` `N_MAX = 100;` `    ``private` `const` `int` `W_MAX = 10000000;`   `    ``public` `static` `int` `MaxWeight(``int``[] w, ``int``[] val, ``int` `n, ``int` `c)` `    ``{` `        ``// Initialize dp array` `        ``int``[,] dp = ``new` `int``[V_SUM_MAX + 1, N_MAX + 1];` `        ``for` `(``int` `i = 0; i <= V_SUM_MAX; i++)` `        ``{` `            ``for` `(``int` `j = 0; j <= n; j++)` `            ``{` `                ``dp[i, j] = W_MAX;` `            ``}` `        ``}`   `        ``// Base case initialization` `        ``for` `(``int` `i = 0; i <= n; i++)` `        ``{` `            ``dp[0, i] = 0;` `        ``}`   `        ``// iterate over subproblems ans get` `        ``// the current value from previous computation` `        ``for` `(``int` `i = 1; i <= V_SUM_MAX; i++)` `        ``{` `            ``for` `(``int` `j = 1; j <= n; j++)` `            ``{` `                ``dp[i, j] = Math.Min(` `                    ``dp[i, j - 1],` `                    ``(i >= val[j - 1])` `                        ``? w[j - 1] + dp[i - val[j - 1], j - 1]` `                        ``: W_MAX);` `            ``}` `        ``}`   `        ``// Finding maximum value` `        ``for` `(``int` `i = V_SUM_MAX; i >= 0; i--)` `        ``{` `            ``if` `(dp[i, n] <= c)` `            ``{` `                ``return` `i;` `            ``}` `        ``}`   `        ``return` `0;` `    ``}`   `    ``public` `static` `void` `Main()` `    ``{` `        ``int``[] w = { 3, 4, 5 };` `        ``int``[] val = { 30, 50, 60 };` `        ``int` `n = w.Length;` `        ``int` `c = 8;`   `        ``Console.WriteLine(MaxWeight(w, val, n, c));` `    ``}` `}`

## Javascript

 `function` `MaxWeight(w, val, n, c) {` `  ``// Initialize dp array` `  ``const V_SUM_MAX = 1000;` `  ``const N_MAX = 100;` `  ``const W_MAX = 10000000;` `  ``const dp = ``new` `Array(V_SUM_MAX + 1).fill().map(() => ``new` `Array(N_MAX + 1).fill(W_MAX));`   `  ``// Base case initialization` `  ``for` `(let i = 0; i <= n; i++) {` `    ``dp[0][i] = 0;` `  ``}`   `  ``// iterate over subproblems and get` `  ``// the current value from previous computation` `  ``for` `(let i = 1; i <= V_SUM_MAX; i++) {` `    ``for` `(let j = 1; j <= n; j++) {` `      ``dp[i][j] = Math.min(` `        ``dp[i][j - 1],` `        ``(i >= val[j - 1]) ? w[j - 1] + dp[i - val[j - 1]][j - 1] : W_MAX` `      ``);` `    ``}` `  ``}`   `  ``// Finding maximum value` `  ``for` `(let i = V_SUM_MAX; i >= 0; i--) {` `    ``if` `(dp[i][n] <= c) {` `      ``return` `i;` `    ``}` `  ``}`   `  ``return` `0;` `}`   `const w = [3, 4, 5];` `const val = [30, 50, 60];` `const n = w.length;` `const c = 8;`   `console.log(MaxWeight(w, val, n, c));`

Output

`90 `

Time Complexity: O(V_sum * N).
Auxiliary Space : O(V_sum * N).

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