# 0/1 Knapsack Problem

Given N items where each item has some weight and profit associated with it and also given a bag with capacity W, [i.e., the bag can hold at most W weight in it]. The task is to put the items into the bag such that the sum of profits associated with them is the maximum possible.

Note: The constraint here is we can either put an item completely into the bag or cannot put it at all [It is not possible to put a part of an item into the bag].

Examples:

Input: N = 3, W = 4, profit[] = {1, 2, 3}, weight[] = {4, 5, 1}
Output: 3
Explanation: There are two items which have weight less than or equal to 4. If we select the item with weight 4, the possible profit is 1. And if we select the item with weight 1, the possible profit is 3. So the maximum possible profit is 3. Note that we cannot put both the items with weight 4 and 1 together as the capacity of the bag is 4.

Input: N = 3, W = 3, profit[] = {1, 2, 3}, weight[] = {4, 5, 6}
Output: 0

Recommended Practice

## Recursion Approach for 0/1 Knapsack Problem:

To solve the problem follow the below idea:

A simple solution is to consider all subsets of items and calculate the total weight and profit of all subsets. Consider the only subsets whose total weight is smaller than W. From all such subsets, pick the subset with maximum profit.

Optimal Substructure: To consider all subsets of items, there can be two cases for every item.

• Case 1: The item is included in the optimal subset.
• Case 2: The item is not included in the optimal set.

Follow the below steps to solve the problem:

The maximum value obtained from ‘N’ items is the max of the following two values.

• Case 1 (include the Nth item): Value of the Nth item plus maximum value obtained by remaining N-1 items and remaining weight i.e. (W-weight of the Nth item).
• Case 2 (exclude the Nth item): Maximum value obtained by N-1 items and W weight.
• If the weight of the ‘Nth‘ item is greater than ‘W’, then the Nth item cannot be included and Case 2 is the only possibility.

Below is the implementation of the above approach:

## C++

 `/* A Naive recursive implementation of ` ` ``0-1 Knapsack problem */` `#include ` `using` `namespace` `std; ` ` `  `// A utility function that returns ` `// maximum of two integers ` `int` `max(``int` `a, ``int` `b) { ``return` `(a > b) ? a : b; } ` ` `  `// Returns the maximum value that ` `// can be put in a knapsack of capacity W ` `int` `knapSack(``int` `W, ``int` `wt[], ``int` `val[], ``int` `n) ` `{ ` ` `  `    ``// Base Case ` `    ``if` `(n == 0 || W == 0) ` `        ``return` `0; ` ` `  `    ``// If weight of the nth item is more ` `    ``// than Knapsack capacity W, then ` `    ``// this item cannot be included ` `    ``// in the optimal solution ` `    ``if` `(wt[n - 1] > W) ` `        ``return` `knapSack(W, wt, val, n - 1); ` ` `  `    ``// Return the maximum of two cases: ` `    ``// (1) nth item included ` `    ``// (2) not included ` `    ``else` `        ``return` `max( ` `            ``val[n - 1] ` `                ``+ knapSack(W - wt[n - 1], wt, val, n - 1), ` `            ``knapSack(W, wt, val, n - 1)); ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `profit[] = { 60, 100, 120 }; ` `    ``int` `weight[] = { 10, 20, 30 }; ` `    ``int` `W = 50; ` `    ``int` `n = ``sizeof``(profit) / ``sizeof``(profit); ` `    ``cout << knapSack(W, weight, profit, n); ` `    ``return` `0; ` `} ` ` `  `// This code is contributed by rathbhupendra`

## C

 `/* A Naive recursive implementation ` `of 0-1 Knapsack problem */` `#include ` ` `  `// A utility function that returns ` `// maximum of two integers ` `int` `max(``int` `a, ``int` `b) { ``return` `(a > b) ? a : b; } ` ` `  `// Returns the maximum value that can be ` `// put in a knapsack of capacity W ` `int` `knapSack(``int` `W, ``int` `wt[], ``int` `val[], ``int` `n) ` `{ ` `    ``// Base Case ` `    ``if` `(n == 0 || W == 0) ` `        ``return` `0; ` ` `  `    ``// If weight of the nth item is more than ` `    ``// Knapsack capacity W, then this item cannot ` `    ``// be included in the optimal solution ` `    ``if` `(wt[n - 1] > W) ` `        ``return` `knapSack(W, wt, val, n - 1); ` ` `  `    ``// Return the maximum of two cases: ` `    ``// (1) nth item included ` `    ``// (2) not included ` `    ``else` `        ``return` `max( ` `            ``val[n - 1] ` `                ``+ knapSack(W - wt[n - 1], wt, val, n - 1), ` `            ``knapSack(W, wt, val, n - 1)); ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `profit[] = { 60, 100, 120 }; ` `    ``int` `weight[] = { 10, 20, 30 }; ` `    ``int` `W = 50; ` `    ``int` `n = ``sizeof``(profit) / ``sizeof``(profit); ` `    ``printf``(``"%d"``, knapSack(W, weight, profit, n)); ` `    ``return` `0; ` `}`

## Java

 `/* A Naive recursive implementation ` `of 0-1 Knapsack problem */` `class` `Knapsack { ` ` `  `    ``// A utility function that returns ` `    ``// maximum of two integers ` `    ``static` `int` `max(``int` `a, ``int` `b) { ``return` `(a > b) ? a : b; } ` ` `  `    ``// Returns the maximum value that ` `    ``// can be put in a knapsack of ` `    ``// capacity W ` `    ``static` `int` `knapSack(``int` `W, ``int` `wt[], ``int` `val[], ``int` `n) ` `    ``{ ` `        ``// Base Case ` `        ``if` `(n == ``0` `|| W == ``0``) ` `            ``return` `0``; ` ` `  `        ``// If weight of the nth item is ` `        ``// more than Knapsack capacity W, ` `        ``// then this item cannot be included ` `        ``// in the optimal solution ` `        ``if` `(wt[n - ``1``] > W) ` `            ``return` `knapSack(W, wt, val, n - ``1``); ` ` `  `        ``// Return the maximum of two cases: ` `        ``// (1) nth item included ` `        ``// (2) not included ` `        ``else` `            ``return` `max(val[n - ``1``] ` `                           ``+ knapSack(W - wt[n - ``1``], wt, ` `                                      ``val, n - ``1``), ` `                       ``knapSack(W, wt, val, n - ``1``)); ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String args[]) ` `    ``{ ` `        ``int` `profit[] = ``new` `int``[] { ``60``, ``100``, ``120` `}; ` `        ``int` `weight[] = ``new` `int``[] { ``10``, ``20``, ``30` `}; ` `        ``int` `W = ``50``; ` `        ``int` `n = profit.length; ` `        ``System.out.println(knapSack(W, weight, profit, n)); ` `    ``} ` `} ` `/*This code is contributed by Rajat Mishra */`

## Python

 `# A naive recursive implementation ` `# of 0-1 Knapsack Problem ` ` `  `# Returns the maximum value that ` `# can be put in a knapsack of ` `# capacity W ` ` `  ` `  `def` `knapSack(W, wt, val, n): ` ` `  `    ``# Base Case ` `    ``if` `n ``=``=` `0` `or` `W ``=``=` `0``: ` `        ``return` `0` ` `  `    ``# If weight of the nth item is ` `    ``# more than Knapsack of capacity W, ` `    ``# then this item cannot be included ` `    ``# in the optimal solution ` `    ``if` `(wt[n``-``1``] > W): ` `        ``return` `knapSack(W, wt, val, n``-``1``) ` ` `  `    ``# return the maximum of two cases: ` `    ``# (1) nth item included ` `    ``# (2) not included ` `    ``else``: ` `        ``return` `max``( ` `            ``val[n``-``1``] ``+` `knapSack( ` `                ``W``-``wt[n``-``1``], wt, val, n``-``1``), ` `            ``knapSack(W, wt, val, n``-``1``)) ` ` `  `# end of function knapSack ` ` `  ` `  `# Driver Code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``profit ``=` `[``60``, ``100``, ``120``] ` `    ``weight ``=` `[``10``, ``20``, ``30``] ` `    ``W ``=` `50` `    ``n ``=` `len``(profit) ` `    ``print` `knapSack(W, weight, profit, n) ` ` `  `# This code is contributed by Nikhil Kumar Singh `

## C#

 `/* A Naive recursive implementation of ` `0-1 Knapsack problem */` `using` `System; ` ` `  `class` `GFG { ` ` `  `    ``// A utility function that returns ` `    ``// maximum of two integers ` `    ``static` `int` `max(``int` `a, ``int` `b) { ``return` `(a > b) ? a : b; } ` ` `  `    ``// Returns the maximum value that can ` `    ``// be put in a knapsack of capacity W ` `    ``static` `int` `knapSack(``int` `W, ``int``[] wt, ``int``[] val, ``int` `n) ` `    ``{ ` ` `  `        ``// Base Case ` `        ``if` `(n == 0 || W == 0) ` `            ``return` `0; ` ` `  `        ``// If weight of the nth item is ` `        ``// more than Knapsack capacity W, ` `        ``// then this item cannot be ` `        ``// included in the optimal solution ` `        ``if` `(wt[n - 1] > W) ` `            ``return` `knapSack(W, wt, val, n - 1); ` ` `  `        ``// Return the maximum of two cases: ` `        ``// (1) nth item included ` `        ``// (2) not included ` `        ``else` `            ``return` `max(val[n - 1] ` `                           ``+ knapSack(W - wt[n - 1], wt, ` `                                      ``val, n - 1), ` `                       ``knapSack(W, wt, val, n - 1)); ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``int``[] profit = ``new` `int``[] { 60, 100, 120 }; ` `        ``int``[] weight = ``new` `int``[] { 10, 20, 30 }; ` `        ``int` `W = 50; ` `        ``int` `n = profit.Length; ` ` `  `        ``Console.WriteLine(knapSack(W, weight, profit, n)); ` `    ``} ` `} ` ` `  `// This code is contributed by Sam007`

## Javascript

 ``

## PHP

 ` ``\$W``) ` `        ``return` `knapSack(``\$W``, ``\$wt``, ``\$val``, ``\$n` `- 1); ` `     `  `    ``// Return the maximum of two cases:  ` `    ``// (1) nth item included  ` `    ``// (2) not included ` `    ``else` `        ``return` `max(``\$val``[``\$n` `- 1] +  ` `               ``knapSack(``\$W` `- ``\$wt``[``\$n` `- 1],  ` `               ``\$wt``, ``\$val``, ``\$n` `- 1),  ` `               ``knapSack(``\$W``, ``\$wt``, ``\$val``, ``\$n``-1)); ` `} ` ` `  `    ``// Driver Code ` `    ``\$profit` `= ``array``(60, 100, 120); ` `    ``\$weight` `= ``array``(10, 20, 30); ` `    ``\$W` `= 50; ` `    ``\$n` `= ``count``(``\$profit``); ` `    ``echo` `knapSack(``\$W``, ``\$weight``, ``\$profit``, ``\$n``); ` ` `  `// This code is contributed by Sam007 ` `?>`

Output

```220
```

Time Complexity: O(2N)
Auxiliary Space: O(N), Stack space required for recursion

## Dynamic Programming Approach for 0/1 Knapsack Problem

### Memoization Approach for 0/1 Knapsack Problem:

Note: It should be noted that the above function using recursion computes the same subproblems again and again. See the following recursion tree, K(1, 1) is being evaluated twice.

In the following recursion tree, K() refers  to knapSack(). The two parameters indicated in the following recursion tree are n and W.
The recursion tree is for following sample inputs.
weight[] = {1, 1, 1}, W = 2, profit[] = {10, 20, 30}

K(3, 2)
/                         \
/                             \
K(2, 2)                                K(2, 1)
/               \                           /               \
/                   \                       /                   \
K(1, 2)               K(1, 1)            K(1, 1)       K(1, 0)
/     \                  /       \                /                  \
/         \              /           \            /                      \
K(0, 2)  K(0, 1)  K(0, 1)  K(0, 0)  K(0, 1)          K(0, 0)

Recursion tree for Knapsack capacity 2 units and 3 items of 1 unit weight.

As there are repetitions of the same subproblem again and again we can implement the following idea to solve the problem.

If we get a subproblem the first time, we can solve this problem by creating a 2-D array that can store a particular state (n, w). Now if we come across the same state (n, w) again instead of calculating it in exponential complexity we can directly return its result stored in the table in constant time.

Below is the implementation of the above approach:

## C++

 `// Here is the top-down approach of ` `// dynamic programming ` `#include ` `using` `namespace` `std; ` ` `  `// Returns the value of maximum profit ` `int` `knapSackRec(``int` `W, ``int` `wt[], ``int` `val[], ``int` `index, ``int``** dp) ` `{ ` `    ``// base condition ` `    ``if` `(index < 0) ` `        ``return` `0; ` `    ``if` `(dp[index][W] != -1) ` `        ``return` `dp[index][W]; ` ` `  `    ``if` `(wt[index] > W) { ` ` `  `        ``// Store the value of function call ` `        ``// stack in table before return ` `        ``dp[index][W] = knapSackRec(W, wt, val, index - 1, dp); ` `        ``return` `dp[index][W]; ` `    ``} ` `    ``else` `{ ` `        ``// Store value in a table before return ` `        ``dp[index][W] = max(val[index] ` `                           ``+ knapSackRec(W - wt[index], wt, val, ` `                                         ``index - 1, dp), ` `                       ``knapSackRec(W, wt, val, index - 1, dp)); ` ` `  `        ``// Return value of table after storing ` `        ``return` `dp[index][W]; ` `    ``} ` `} ` ` `  `int` `knapSack(``int` `W, ``int` `wt[], ``int` `val[], ``int` `n) ` `{ ` `    ``// double pointer to declare the ` `    ``// table dynamically ` `    ``int``** dp; ` `    ``dp = ``new` `int``*[n]; ` ` `  `    ``// loop to create the table dynamically ` `    ``for` `(``int` `i = 0; i < n; i++) ` `        ``dp[i] = ``new` `int``[W + 1]; ` ` `  `    ``// loop to initially filled the ` `    ``// table with -1 ` `    ``for` `(``int` `i = 0; i < n; i++) ` `        ``for` `(``int` `j = 0; j < W + 1; j++) ` `            ``dp[i][j] = -1; ` `    ``return` `knapSackRec(W, wt, val, n - 1, dp); ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `profit[] = { 60, 100, 120 }; ` `    ``int` `weight[] = { 10, 20, 30 }; ` `    ``int` `W = 50; ` `    ``int` `n = ``sizeof``(profit) / ``sizeof``(profit); ` `    ``cout << knapSack(W, weight, profit, n); ` `    ``return` `0; ` `}`

## Java

 `// Here is the top-down approach of ` `// dynamic programming ` ` `  `import` `java.io.*; ` ` `  `class` `GFG { ` ` `  `    ``// A utility function that returns ` `    ``// maximum of two integers ` `    ``static` `int` `max(``int` `a, ``int` `b) { ``return` `(a > b) ? a : b; } ` ` `  `    ``// Returns the value of maximum profit ` `    ``static` `int` `knapSackRec(``int` `W, ``int` `wt[], ``int` `val[], ` `                           ``int` `n, ``int``[][] dp) ` `    ``{ ` ` `  `        ``// Base condition ` `        ``if` `(n == ``0` `|| W == ``0``) ` `            ``return` `0``; ` ` `  `        ``if` `(dp[n][W] != -``1``) ` `            ``return` `dp[n][W]; ` ` `  `        ``if` `(wt[n - ``1``] > W) ` ` `  `            ``// Store the value of function call ` `            ``// stack in table before return ` `            ``return` `dp[n][W] ` `                ``= knapSackRec(W, wt, val, n - ``1``, dp); ` ` `  `        ``else` ` `  `            ``// Return value of table after storing ` `            ``return` `dp[n][W] ` `                ``= max((val[n - ``1``] ` `                       ``+ knapSackRec(W - wt[n - ``1``], wt, val, ` `                                     ``n - ``1``, dp)), ` `                      ``knapSackRec(W, wt, val, n - ``1``, dp)); ` `    ``} ` ` `  `    ``static` `int` `knapSack(``int` `W, ``int` `wt[], ``int` `val[], ``int` `N) ` `    ``{ ` ` `  `        ``// Declare the table dynamically ` `        ``int` `dp[][] = ``new` `int``[N + ``1``][W + ``1``]; ` ` `  `        ``// Loop to initially filled the ` `        ``// table with -1 ` `        ``for` `(``int` `i = ``0``; i < N + ``1``; i++) ` `            ``for` `(``int` `j = ``0``; j < W + ``1``; j++) ` `                ``dp[i][j] = -``1``; ` ` `  `        ``return` `knapSackRec(W, wt, val, N, dp); ` `    ``} ` ` `  `    ``// Driver Code ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``int` `profit[] = { ``60``, ``100``, ``120` `}; ` `        ``int` `weight[] = { ``10``, ``20``, ``30` `}; ` ` `  `        ``int` `W = ``50``; ` `        ``int` `N = profit.length; ` ` `  `        ``System.out.println(knapSack(W, weight, profit, N)); ` `    ``} ` `} ` ` `  `// This Code is contributed By FARAZ AHMAD`

## Python3

 `# This is the memoization approach of ` `# 0 / 1 Knapsack in Python in simple ` `# we can say recursion + memoization = DP ` ` `  ` `  `def` `knapsack(wt, val, W, n): ` ` `  `    ``# base conditions ` `    ``if` `n ``=``=` `0` `or` `W ``=``=` `0``: ` `        ``return` `0` `    ``if` `t[n][W] !``=` `-``1``: ` `        ``return` `t[n][W] ` ` `  `    ``# choice diagram code ` `    ``if` `wt[n``-``1``] <``=` `W: ` `        ``t[n][W] ``=` `max``( ` `            ``val[n``-``1``] ``+` `knapsack( ` `                ``wt, val, W``-``wt[n``-``1``], n``-``1``), ` `            ``knapsack(wt, val, W, n``-``1``)) ` `        ``return` `t[n][W] ` `    ``elif` `wt[n``-``1``] > W: ` `        ``t[n][W] ``=` `knapsack(wt, val, W, n``-``1``) ` `        ``return` `t[n][W] ` ` `  `# Driver code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``profit ``=` `[``60``, ``100``, ``120``] ` `    ``weight ``=` `[``10``, ``20``, ``30``] ` `    ``W ``=` `50` `    ``n ``=` `len``(profit) ` `     `  `    ``# We initialize the matrix with -1 at first. ` `    ``t ``=` `[[``-``1` `for` `i ``in` `range``(W ``+` `1``)] ``for` `j ``in` `range``(n ``+` `1``)] ` `    ``print``(knapsack(weight, profit, W, n)) ` ` `  `# This code is contributed by Prosun Kumar Sarkar `

## C#

 `// Here is the top-down approach of ` `// dynamic programming ` `using` `System; ` `public` `class` `GFG { ` ` `  `    ``// A utility function that returns ` `    ``// maximum of two integers ` `    ``static` `int` `max(``int` `a, ``int` `b) { ``return` `(a > b) ? a : b; } ` ` `  `    ``// Returns the value of maximum profit ` `    ``static` `int` `knapSackRec(``int` `W, ``int``[] wt, ``int``[] val, ` `                           ``int` `n, ``int``[, ] dp) ` `    ``{ ` ` `  `        ``// Base condition ` `        ``if` `(n == 0 || W == 0) ` `            ``return` `0; ` `        ``if` `(dp[n, W] != -1) ` `            ``return` `dp[n, W]; ` `        ``if` `(wt[n - 1] > W) ` ` `  `            ``// Store the value of function call ` `            ``// stack in table before return ` `            ``return` `dp[n, W] ` `                ``= knapSackRec(W, wt, val, n - 1, dp); ` ` `  `        ``else` ` `  `            ``// Return value of table after storing ` `            ``return` `dp[n, W] ` `                ``= max((val[n - 1] ` `                       ``+ knapSackRec(W - wt[n - 1], wt, val, ` `                                     ``n - 1, dp)), ` `                      ``knapSackRec(W, wt, val, n - 1, dp)); ` `    ``} ` ` `  `    ``static` `int` `knapSack(``int` `W, ``int``[] wt, ``int``[] val, ``int` `N) ` `    ``{ ` ` `  `        ``// Declare the table dynamically ` `        ``int``[, ] dp = ``new` `int``[N + 1, W + 1]; ` ` `  `        ``// Loop to initially filled the ` `        ``// table with -1 ` `        ``for` `(``int` `i = 0; i < N + 1; i++) ` `            ``for` `(``int` `j = 0; j < W + 1; j++) ` `                ``dp[i, j] = -1; ` ` `  `        ``return` `knapSackRec(W, wt, val, N, dp); ` `    ``} ` ` `  `    ``// Driver Code ` `    ``static` `public` `void` `Main() ` `    ``{ ` ` `  `        ``int``[] profit = ``new` `int``[] { 60, 100, 120 }; ` `        ``int``[] weight = ``new` `int``[] { 10, 20, 30 }; ` ` `  `        ``int` `W = 50; ` `        ``int` `N = profit.Length; ` ` `  `        ``Console.WriteLine(knapSack(W, weight, profit, N)); ` `    ``} ` `} ` ` `  `// This Code is contributed By Dharanendra L V.`

## Javascript

 ``

Output

```220
```

Time Complexity: O(N * W). As redundant calculations of states are avoided.
Auxiliary Space: O(N * W) + O(N). The use of a 2D array data structure for storing intermediate states and O(N) auxiliary stack space(ASS) has been used for recursion stack

### Bottom-up Approach for 0/1 Knapsack Problem:

To solve the problem follow the below idea:

Since subproblems are evaluated again, this problem has Overlapping Sub-problems property. So the 0/1 Knapsack problem has both properties (see this and this) of a dynamic programming problem. Like other typical Dynamic Programming(DP) problems, re-computation of the same subproblems can be avoided by constructing a temporary array K[][] in a bottom-up manner.

Illustration:

Below is the illustration of the above approach:

Let, weight[] = {1, 2, 3}, profit[] = {10, 15, 40}, Capacity = 6

• If no element is filled, then the possible profit is 0.
weight⇢
item⇣/
0 1 2 3 4 5 6
0 0 0 0 0 0 0 0
1
2
3
• For filling the first item in the bag: If we follow the above mentioned procedure, the table will look like the following.
weight⇢
item⇣/
0 1 2 3 4 5 6
0 0 0 0 0 0 0 0
1 0 10 10 10 10 10 10
2
3
• For filling the second item:
When jthWeight = 2, then maximum possible profit is max (10, DP[2-2] + 15) = max(10, 15) = 15.
When jthWeight = 3, then maximum possible profit is  max(2 not put, 2 is put into bag) = max(DP, 15+DP[3-2]) = max(10, 25) = 25.
weight⇢
item⇣/
0 1 2 3 4 5 6
0 0 0 0 0 0 0 0
1 0 10 10 10 10 10 10
2 0 10 15 25 25 25 25
3
• For filling the third item:
When jthWeight = 3, the maximum possible profit is max(DP, 40+DP[3-3]) = max(25, 40) = 40.
When jthWeight = 4, the maximum possible profit is max(DP, 40+DP[4-3]) = max(25, 50) = 50.
When jthWeight = 5, the maximum possible profit is max(DP, 40+DP[5-3]) = max(25, 55) = 55.
When jthWeight = 6, the maximum possible profit is max(DP, 40+DP[6-3]) = max(25, 65) = 65.
weight⇢
item⇣/
0 1 2 3 4 5 6
0 0 0 0 0 0 0 0
1 0 10 10 10 10 10 10
2 0 10 15 25 25 25 25
3 0 10 15 40 50 55 65

Below is the implementation of the above approach:

## C++

 `// A dynamic programming based ` `// solution for 0-1 Knapsack problem ` `#include ` `using` `namespace` `std; ` ` `  `// A utility function that returns ` `// maximum of two integers ` `int` `max(``int` `a, ``int` `b) { ``return` `(a > b) ? a : b; } ` ` `  `// Returns the maximum value that ` `// can be put in a knapsack of capacity W ` `int` `knapSack(``int` `W, ``int` `wt[], ``int` `val[], ``int` `n) ` `{ ` `    ``int` `i, w; ` `    ``vector > K(n + 1, vector<``int``>(W + 1)); ` ` `  `    ``// Build table K[][] in bottom up manner ` `    ``for` `(i = 0; i <= n; i++) { ` `        ``for` `(w = 0; w <= W; w++) { ` `            ``if` `(i == 0 || w == 0) ` `                ``K[i][w] = 0; ` `            ``else` `if` `(wt[i - 1] <= w) ` `                ``K[i][w] = max(val[i - 1] ` `                                  ``+ K[i - 1][w - wt[i - 1]], ` `                              ``K[i - 1][w]); ` `            ``else` `                ``K[i][w] = K[i - 1][w]; ` `        ``} ` `    ``} ` `    ``return` `K[n][W]; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `profit[] = { 60, 100, 120 }; ` `    ``int` `weight[] = { 10, 20, 30 }; ` `    ``int` `W = 50; ` `    ``int` `n = ``sizeof``(profit) / ``sizeof``(profit); ` ` `  `    ``cout << knapSack(W, weight, profit, n); ` ` `  `    ``return` `0; ` `} ` ` `  `// This code is contributed by Debojyoti Mandal`

## C

 `// A Dynamic Programming based ` `// solution for 0-1 Knapsack problem ` `#include ` ` `  `// A utility function that returns ` `// maximum of two integers ` `int` `max(``int` `a, ``int` `b) { ``return` `(a > b) ? a : b; } ` ` `  `// Returns the maximum value that ` `// can be put in a knapsack of capacity W ` `int` `knapSack(``int` `W, ``int` `wt[], ``int` `val[], ``int` `n) ` `{ ` `    ``int` `i, w; ` `    ``int` `K[n + 1][W + 1]; ` ` `  `    ``// Build table K[][] in bottom up manner ` `    ``for` `(i = 0; i <= n; i++) { ` `        ``for` `(w = 0; w <= W; w++) { ` `            ``if` `(i == 0 || w == 0) ` `                ``K[i][w] = 0; ` `            ``else` `if` `(wt[i - 1] <= w) ` `                ``K[i][w] = max(val[i - 1] ` `                                  ``+ K[i - 1][w - wt[i - 1]], ` `                              ``K[i - 1][w]); ` `            ``else` `                ``K[i][w] = K[i - 1][w]; ` `        ``} ` `    ``} ` ` `  `    ``return` `K[n][W]; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `profit[] = { 60, 100, 120 }; ` `    ``int` `weight[] = { 10, 20, 30 }; ` `    ``int` `W = 50; ` `    ``int` `n = ``sizeof``(profit) / ``sizeof``(profit); ` `    ``printf``(``"%d"``, knapSack(W, weight, profit, n)); ` `    ``return` `0; ` `}`

## Java

 `// A Dynamic Programming based solution ` `// for 0-1 Knapsack problem ` ` `  `import` `java.io.*; ` ` `  `class` `Knapsack { ` ` `  `    ``// A utility function that returns ` `    ``// maximum of two integers ` `    ``static` `int` `max(``int` `a, ``int` `b) { ``return` `(a > b) ? a : b; } ` ` `  `    ``// Returns the maximum value that can ` `    ``// be put in a knapsack of capacity W ` `    ``static` `int` `knapSack(``int` `W, ``int` `wt[], ``int` `val[], ``int` `n) ` `    ``{ ` `        ``int` `i, w; ` `        ``int` `K[][] = ``new` `int``[n + ``1``][W + ``1``]; ` ` `  `        ``// Build table K[][] in bottom up manner ` `        ``for` `(i = ``0``; i <= n; i++) { ` `            ``for` `(w = ``0``; w <= W; w++) { ` `                ``if` `(i == ``0` `|| w == ``0``) ` `                    ``K[i][w] = ``0``; ` `                ``else` `if` `(wt[i - ``1``] <= w) ` `                    ``K[i][w] ` `                        ``= max(val[i - ``1``] ` `                                  ``+ K[i - ``1``][w - wt[i - ``1``]], ` `                              ``K[i - ``1``][w]); ` `                ``else` `                    ``K[i][w] = K[i - ``1``][w]; ` `            ``} ` `        ``} ` ` `  `        ``return` `K[n][W]; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String args[]) ` `    ``{ ` `        ``int` `profit[] = ``new` `int``[] { ``60``, ``100``, ``120` `}; ` `        ``int` `weight[] = ``new` `int``[] { ``10``, ``20``, ``30` `}; ` `        ``int` `W = ``50``; ` `        ``int` `n = profit.length; ` `        ``System.out.println(knapSack(W, weight, profit, n)); ` `    ``} ` `} ` `/*This code is contributed by Rajat Mishra */`

## Python3

 `# A Dynamic Programming based Python ` `# Program for 0-1 Knapsack problem ` `# Returns the maximum value that can ` `# be put in a knapsack of capacity W ` ` `  ` `  `def` `knapSack(W, wt, val, n): ` `    ``K ``=` `[[``0` `for` `x ``in` `range``(W ``+` `1``)] ``for` `x ``in` `range``(n ``+` `1``)] ` ` `  `    ``# Build table K[][] in bottom up manner ` `    ``for` `i ``in` `range``(n ``+` `1``): ` `        ``for` `w ``in` `range``(W ``+` `1``): ` `            ``if` `i ``=``=` `0` `or` `w ``=``=` `0``: ` `                ``K[i][w] ``=` `0` `            ``elif` `wt[i``-``1``] <``=` `w: ` `                ``K[i][w] ``=` `max``(val[i``-``1``] ` `                              ``+` `K[i``-``1``][w``-``wt[i``-``1``]], ` `                              ``K[i``-``1``][w]) ` `            ``else``: ` `                ``K[i][w] ``=` `K[i``-``1``][w] ` ` `  `    ``return` `K[n][W] ` ` `  ` `  `# Driver code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``profit ``=` `[``60``, ``100``, ``120``] ` `    ``weight ``=` `[``10``, ``20``, ``30``] ` `    ``W ``=` `50` `    ``n ``=` `len``(profit) ` `    ``print``(knapSack(W, weight, profit, n)) ` ` `  `# This code is contributed by Bhavya Jain `

## C#

 `// A Dynamic Programming based solution for ` `// 0-1 Knapsack problem ` `using` `System; ` ` `  `class` `GFG { ` ` `  `    ``// A utility function that returns ` `    ``// maximum of two integers ` `    ``static` `int` `max(``int` `a, ``int` `b) { ``return` `(a > b) ? a : b; } ` ` `  `    ``// Returns the maximum value that ` `    ``// can be put in a knapsack of ` `    ``// capacity W ` `    ``static` `int` `knapSack(``int` `W, ``int``[] wt, ``int``[] val, ``int` `n) ` `    ``{ ` `        ``int` `i, w; ` `        ``int``[, ] K = ``new` `int``[n + 1, W + 1]; ` ` `  `        ``// Build table K[][] in bottom ` `        ``// up manner ` `        ``for` `(i = 0; i <= n; i++) { ` `            ``for` `(w = 0; w <= W; w++) { ` `                ``if` `(i == 0 || w == 0) ` `                    ``K[i, w] = 0; ` ` `  `                ``else` `if` `(wt[i - 1] <= w) ` `                    ``K[i, w] = Math.Max( ` `                        ``val[i - 1] ` `                            ``+ K[i - 1, w - wt[i - 1]], ` `                        ``K[i - 1, w]); ` `                ``else` `                    ``K[i, w] = K[i - 1, w]; ` `            ``} ` `        ``} ` ` `  `        ``return` `K[n, W]; ` `    ``} ` ` `  `    ``// Driver code ` `    ``static` `void` `Main() ` `    ``{ ` `        ``int``[] profit = ``new` `int``[] { 60, 100, 120 }; ` `        ``int``[] weight = ``new` `int``[] { 10, 20, 30 }; ` `        ``int` `W = 50; ` `        ``int` `n = profit.Length; ` ` `  `        ``Console.WriteLine(knapSack(W, weight, profit, n)); ` `    ``} ` `} ` ` `  `// This code is contributed by Sam007`

## Javascript

 ``

## PHP

 ``

Output

```220
```

Time Complexity: O(N * W). where ‘N’ is the number of elements and ‘W’ is capacity.
Auxiliary Space: O(N * W). The use of a 2-D array of size ‘N*W’.

### Space optimized Approach for 0/1 Knapsack Problem using Dynamic Programming:

To solve the problem follow the below idea:

For calculating the current row of the dp[] array we require only previous row, but if we start traversing the rows from right to left then it can be done with a single row only

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Function to find the maximum profit ` `int` `knapSack(``int` `W, ``int` `wt[], ``int` `val[], ``int` `n) ` `{ ` `    ``// Making and initializing dp array ` `    ``int` `dp[W + 1]; ` `    ``memset``(dp, 0, ``sizeof``(dp)); ` ` `  `    ``for` `(``int` `i = 1; i < n + 1; i++) { ` `        ``for` `(``int` `w = W; w >= 0; w--) { ` ` `  `            ``if` `(wt[i - 1] <= w) ` `                 `  `                ``// Finding the maximum value ` `                ``dp[w] = max(dp[w], ` `                            ``dp[w - wt[i - 1]] + val[i - 1]); ` `        ``} ` `    ``} ` `    ``// Returning the maximum value of knapsack ` `    ``return` `dp[W]; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `profit[] = { 60, 100, 120 }; ` `    ``int` `weight[] = { 10, 20, 30 }; ` `    ``int` `W = 50; ` `    ``int` `n = ``sizeof``(profit) / ``sizeof``(profit); ` `    ``cout << knapSack(W, weight, profit, n); ` `    ``return` `0; ` `}`

## Java

 `// Java program for the above approach ` ` `  `import` `java.util.*; ` ` `  `class` `GFG { ` `    ``static` `int` `knapSack(``int` `W, ``int` `wt[], ``int` `val[], ``int` `n) ` `    ``{ ` `        ``// Making and initializing dp array ` `        ``int``[] dp = ``new` `int``[W + ``1``]; ` ` `  `        ``for` `(``int` `i = ``1``; i < n + ``1``; i++) { ` `            ``for` `(``int` `w = W; w >= ``0``; w--) { ` ` `  `                ``if` `(wt[i - ``1``] <= w) ` ` `  `                    ``// Finding the maximum value ` `                    ``dp[w] ` `                        ``= Math.max(dp[w], dp[w - wt[i - ``1``]] ` `                                              ``+ val[i - ``1``]); ` `            ``} ` `        ``} ` `        ``// Returning the maximum value of knapsack ` `        ``return` `dp[W]; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``int` `profit[] = { ``60``, ``100``, ``120` `}; ` `        ``int` `weight[] = { ``10``, ``20``, ``30` `}; ` `        ``int` `W = ``50``; ` `        ``int` `n = profit.length; ` `        ``System.out.print(knapSack(W, weight, profit, n)); ` `    ``} ` `} ` ` `  `// This code is contributed by gauravrajput1`

## Python3

 `# Python code to implement the above approach ` ` `  ` `  `def` `knapSack(W, wt, val, n): ` `     `  `    ``# Making the dp array ` `    ``dp ``=` `[``0` `for` `i ``in` `range``(W``+``1``)] ` ` `  `    ``# Taking first i elements ` `    ``for` `i ``in` `range``(``1``, n``+``1``): ` `       `  `        ``# Starting from back, ` `        ``# so that we also have data of ` `        ``# previous computation when taking i-1 items ` `        ``for` `w ``in` `range``(W, ``0``, ``-``1``): ` `            ``if` `wt[i``-``1``] <``=` `w: ` `                 `  `                ``# Finding the maximum value ` `                ``dp[w] ``=` `max``(dp[w], dp[w``-``wt[i``-``1``]]``+``val[i``-``1``]) ` `     `  `    ``# Returning the maximum value of knapsack ` `    ``return` `dp[W] ` ` `  ` `  `# Driver code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``profit ``=` `[``60``, ``100``, ``120``] ` `    ``weight ``=` `[``10``, ``20``, ``30``] ` `    ``W ``=` `50` `    ``n ``=` `len``(profit) ` `    ``print``(knapSack(W, weight, profit, n)) ` ` `  ` `  `# This code is contributed by Suyash Saxena`

## C#

 `// Java program for the above approach ` ` `  `using` `System; ` `public` `class` `GFG { ` `    ``static` `int` `knapSack(``int` `W, ``int``[] wt, ``int``[] val, ``int` `n) ` `    ``{ ` `        ``// Making and initializing dp array ` `        ``int``[] dp = ``new` `int``[W + 1]; ` ` `  `        ``for` `(``int` `i = 1; i < n + 1; i++) { ` `            ``for` `(``int` `w = W; w >= 0; w--) { ` ` `  `                ``if` `(wt[i - 1] <= w) ` ` `  `                    ``// Finding the maximum value ` `                    ``dp[w] ` `                        ``= Math.Max(dp[w], dp[w - wt[i - 1]] ` `                                              ``+ val[i - 1]); ` `            ``} ` `        ``} ` `        ``// Returning the maximum value of knapsack ` `        ``return` `dp[W]; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `Main(String[] args) ` `    ``{ ` `        ``int``[] profit = { 60, 100, 120 }; ` `        ``int``[] weight = { 10, 20, 30 }; ` `        ``int` `W = 50; ` `        ``int` `n = profit.Length; ` `        ``Console.Write(knapSack(W, weight, profit, n)); ` `    ``} ` `} ` ` `  `// This code is contributed by gauravrajput1`

## Javascript

 ``

Output

```220
```

Time Complexity: O(N * W). As redundant calculations of states are avoided
Auxiliary Space: O(W) As we are using a 1-D array instead of a 2-D array

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