0/1 Knapsack Problem

Prerequisites: Introduction to Knapsack Problem, its Types and How to solve them

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

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 N^th item): Value of the N^th item plus maximum value obtained by remaining N-1 items and remaining weight i.e. (W-weight of the N^th item).
• Case 2 (exclude the N^th item): Maximum value obtained by N-1 items and W weight.
• If the weight of the ‘N^th‘ 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: 

/* A Naive recursive implementation of
 0-1 Knapsack problem */
#include <bits/stdc++.h>
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[0]);
    cout << knapSack(W, weight, profit, n);
    return 0;
}

// This code is contributed by rathbhupendra

/* A Naive recursive implementation
of 0-1 Knapsack problem */
#include <stdio.h>

// 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[0]);
    printf("%d", knapSack(W, weight, profit, n));
    return 0;
}

/* 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 */

# 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

/* 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

<script>

    /* A Naive recursive implementation of
    0-1 Knapsack problem */
    
    // A utility function that returns
    // maximum of two integers
    function max(a, b)
    {
         return (a > b) ? a : b;
    }
 
    // Returns the maximum value that can
    // be put in a knapsack of capacity W
    function knapSack(W, wt, val, 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));
    }
      
    let profit = [ 60, 100, 120 ];
    let weight = [ 10, 20, 30 ];
    let W = 50;
    let n = profit.length;

    document.write(knapSack(W, weight, profit, n));
    
</script>

<?php
// A Naive recursive implementation
// of 0-1 Knapsack problem 

// Returns the maximum value that
// can be put in a knapsack of 
// capacity W
function knapSack($W, $wt, $val, $n)
{
    // Base Case
    if ($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
    $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
?>

220

Time Complexity: O(2^N)
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:

// Here is the top-down approach of
// dynamic programming
#include <bits/stdc++.h>
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[0]);
    cout << knapSack(W, weight, profit, n);
    return 0;
}

// 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

// 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.

<script>
// A utility function that returns
// maximum of two integers    
function max(a,  b)    
{    
    return (a > b) ? a : b;    
}

// Returns the value of maximum profit
function knapSackRec(W, wt, val, n,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));            
}

function knapSack( W, wt,val,N)
{
    
    // Declare the dp table dynamically 
    // Intializing dp tables(row and cols) with -1 below
     var dp = new Array(N+1).fill(-1).map(el => new Array(W+1).fill(-1));
    return knapSackRec(W, wt, val, N, dp);    
}


   var profit= [ 60, 100, 120 ];
    var weight = [ 10, 20, 30 ];
    
    var W = 50;
    var N = profit.length;        
    
    document.write(knapSack(W, weight, profit, N));

// This code is contributed by akshitsaxenaa09.

</script>

# 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

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[1][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[1][3], 15+DP[1][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[2][3], 40+DP[2][3-3]) = max(25, 40) = 40.
  When jthWeight = 4, the maximum possible profit is max(DP[2][4], 40+DP[2][4-3]) = max(25, 50) = 50.
  When jthWeight = 5, the maximum possible profit is max(DP[2][5], 40+DP[2][5-3]) = max(25, 55) = 55.
  When jthWeight = 6, the maximum possible profit is max(DP[2][6], 40+DP[2][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: 

// A dynamic programming based
// solution for 0-1 Knapsack problem
#include <bits/stdc++.h>
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<vector<int> > 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[0]);

    cout << knapSack(W, weight, profit, n);

    return 0;
}

// This code is contributed by Debojyoti Mandal

// A Dynamic Programming based
// solution for 0-1 Knapsack problem
#include <stdio.h>

// 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[0]);
    printf("%d", knapSack(W, weight, profit, n));
    return 0;
}

// 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 */

// 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

<script>
    // A Dynamic Programming based solution
    // for 0-1 Knapsack problem
    
    // A utility function that returns
    // maximum of two integers
    function max(a, b)
    {
          return (a > b) ? a : b;
    }
 
    // Returns the maximum value that can
    // be put in a knapsack of capacity W
    function knapSack(W, wt, val, n)
    {
        let i, w;
        let K = new Array(n + 1);
 
        // Build table K[][] in bottom up manner
        for (i = 0; i <= n; i++)
        {
            K[i] = new Array(W + 1);
            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];
    }
    
    let profit = [ 60, 100, 120 ];
    let weight = [ 10, 20, 30 ];
    let W = 50;
    let n = profit.length;
    document.write(knapSack(W, weight, profit, n));
</script>

<?php
// A Dynamic Programming based solution
// for 0-1 Knapsack problem

// Returns the maximum value that
// can be put in a knapsack of 
// capacity W
function knapSack($W, $wt, $val, $n)
{
    
    $K = array(array());
    
    // 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
    $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.
?>

# 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

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++ program for the above approach

#include <bits/stdc++.h>
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[0]);
    cout << knapSack(W, weight, profit, n);
    return 0;
}

// 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

// 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