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0-1 Knapsack Problem | DP-10

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  • Difficulty Level : Medium
  • Last Updated : 05 Dec, 2022
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Given weights and values of N items, put these items in a knapsack of capacity W to get the maximum total value in the knapsack. In other words, given two integer arrays val[0..N-1] and wt[0..N-1] which represent values and weights associated with N items respectively. Also given an integer W which represents knapsack capacity, find out the maximum value subset of val[] such that the sum of the weights of this subset is smaller than or equal to W. You cannot break an item, either pick the complete item or don’t pick it (0-1 property)

Example:

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

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

Recommended Practice

0-1 Knapsack Problem using recursion:

To solve the problem follow the below idea:

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

Optimal Sub-structure: 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. 

  • Maximum value obtained by N-1 items and W weight (excluding nth item)
  • Value of nth item plus maximum value obtained by N-1 items and W minus the weight of the Nth item (including Nth item)
  • If the weight of the ‘Nth’ item is greater than ‘W’, then the Nth item cannot be included and Case 1 is the only possibility.

Below is the implementation of the above approach: 

C++




/* 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 val[] = { 60, 100, 120 };
    int wt[] = { 10, 20, 30 };
    int W = 50;
    int n = sizeof(val) / sizeof(val[0]);
    cout << knapSack(W, wt, val, n);
    return 0;
}
 
// This code is contributed by rathbhupendra


C




/* 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 val[] = { 60, 100, 120 };
    int wt[] = { 10, 20, 30 };
    int W = 50;
    int n = sizeof(val) / sizeof(val[0]);
    printf("%d", knapSack(W, wt, val, 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 val[] = new int[] { 60, 100, 120 };
        int wt[] = new int[] { 10, 20, 30 };
        int W = 50;
        int n = val.length;
        System.out.println(knapSack(W, wt, val, 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
val = [60, 100, 120]
wt = [10, 20, 30]
W = 50
n = len(val)
print knapSack(W, wt, val, 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[] val = new int[] { 60, 100, 120 };
        int[] wt = new int[] { 10, 20, 30 };
        int W = 50;
        int n = val.Length;
 
        Console.WriteLine(knapSack(W, wt, val, n));
    }
}
 
// This code is contributed by Sam007


PHP




<?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
    $val = array(60, 100, 120);
    $wt = array(10, 20, 30);
    $W = 50;
    $n = count($val);
    echo knapSack($W, $wt, $val, $n);
 
// This code is contributed by Sam007
?>


Javascript




<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 val = [ 60, 100, 120 ];
    let wt = [ 10, 20, 30 ];
       let W = 50;
    let n = val.length;
 
    document.write(knapSack(W, wt, val, n));
     
</script>


Output

220

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

Note: It should be noted that the above function computes the same sub-problems 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.
wt[] = {1, 1, 1}, W = 2, val[] = {10, 20, 30}

                                         K(n, W)

                                         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.

0-1 Knapsack Problem using memoization:

To solve the problem follow the below idea:

This method is basically an extension to the recursive approach so that we can overcome the problem of calculating redundant cases and thus increased complexity. If we get it 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. This method gives an edge over the recursive approach in this aspect.

Below is the implementation of the above approach:

C++




// 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 i, int** dp)
{
    // base condition
    if (i < 0)
        return 0;
    if (dp[i][W] != -1)
        return dp[i][W];
 
    if (wt[i] > W) {
 
        // Store the value of function call
        // stack in table before return
        dp[i][W] = knapSackRec(W, wt, val, i - 1, dp);
        return dp[i][W];
    }
    else {
        // Store value in a table before return
        dp[i][W] = max(val[i]
                           + knapSackRec(W - wt[i], wt, val,
                                         i - 1, dp),
                       knapSackRec(W, wt, val, i - 1, dp));
 
        // Return value of table after storing
        return dp[i][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 val[] = { 60, 100, 120 };
    int wt[] = { 10, 20, 30 };
    int W = 50;
    int n = sizeof(val) / sizeof(val[0]);
    cout << knapSack(W, wt, val, 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 val[] = { 60, 100, 120 };
        int wt[] = { 10, 20, 30 };
 
        int W = 50;
        int N = val.length;
 
        System.out.println(knapSack(W, wt, val, 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
 
# Driver code
val = [60, 100, 120]
wt = [10, 20, 30]
W = 50
n = len(val)
 
# We initialize the matrix with -1 at first.
t = [[-1 for i in range(W + 1)] for j in range(n + 1)]
 
 
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]
 
 
print(knapsack(wt, val, 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[] val = new int[] { 60, 100, 120 };
        int[] wt = new int[] { 10, 20, 30 };
 
        int W = 50;
        int N = val.Length;
 
        Console.WriteLine(knapSack(W, wt, val, N));
    }
}
 
// This Code is contributed By Dharanendra L V.


Javascript




<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 table dynamically
    var dp = new Array(N + 1);
    for(var i = 0; i < dp.length; i++)
    {
        dp[i]=new Array(W + 1);
      }
     
    // Loop to initially filled the
    // table with -1
    for(var i = 0; i < N + 1; i++)
        for(var j = 0; j < W + 1; j++)
            dp[i][j] = -1;
     
    return knapSackRec(W, wt, val, N, dp);   
}
 
 
   var val= [ 60, 100, 120 ];
    var wt = [ 10, 20, 30 ];
     
    var W = 50;
    var N = val.length;       
     
    document.write(knapSack(W, wt, val, N));
 
// This code is contributed by akshitsaxenaa09.
 
</script>


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

0-1 Knapsack Problem using dynamic programming:

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. 

Follow the below steps to solve the problem:

  • Consider the same cases as mentioned in the recursive approach. 
  • In a DP[][] table let’s consider all the possible weights from ‘1’ to ‘W’ as the columns and weights that can be kept as rows. 
  • The state DP[i][j] will denote the maximum value of ‘j-weight’ considering all values from ‘1 to ith’. So if we consider ‘wi’ (weight in ‘ith’ row) we can fill it in all columns which have ‘weight values > wi’. Now two possibilities can take place: 
    • Fill ‘wi’ in the given column.
    • Do not fill ‘wi’ in the given column.
  • Now we have to take a maximum of these two possibilities, formally if we do not fill the ‘ith’ weight in the ‘jth’ column then the DP[i][j] state will be the same as DP[i-1][j] but if we fill the weight, DP[i][j] will be equal to the value of ‘wi’+ value of the column weighing ‘j-wi’ in the previous row. 
  • So we take the maximum of these two possibilities to fill the current state. 

Below is the illustration of the above approach:

Let weight elements = {1, 2, 3}
Let weight values = {10, 15, 40}
Capacity=6

    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

 Explanation:

For filling ‘weight = 2’ we come 
across ‘j = 3’ in which 
we take maximum of 
(10, 15 + DP[1][3-2]) = 25   
   |                        |
‘2 not filled’     ‘2 filled’

    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

Explanation:

For filling ‘weight=3’, 
we come across ‘j=4’ in which 
we take maximum of (25, 40 + DP[2][4-3])  = 50

For filling ‘weight=3’ 
we come across ‘j=5’ in which 
we take maximum of (25, 40 + DP[2][5-3]) = 55

For filling ‘weight=3’ 
we come across ‘j=6’ in which 
we take maximum of (25, 40 + DP[2][6-3]) = 65

Below is the implementation of the above approach: 

C++




// 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 val[] = { 60, 100, 120 };
    int wt[] = { 10, 20, 30 };
    int W = 50;
    int n = sizeof(val) / sizeof(val[0]);
 
    cout << knapSack(W, wt, val, n);
 
    return 0;
}
 
// This code is contributed by Debojyoti Mandal


C




// 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 val[] = { 60, 100, 120 };
    int wt[] = { 10, 20, 30 };
    int W = 50;
    int n = sizeof(val) / sizeof(val[0]);
    printf("%d", knapSack(W, wt, val, 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 val[] = new int[] { 60, 100, 120 };
        int wt[] = new int[] { 10, 20, 30 };
        int W = 50;
        int n = val.length;
        System.out.println(knapSack(W, wt, val, n));
    }
}
/*This code is contributed by Rajat Mishra */


Python




# 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
val = [60, 100, 120]
wt = [10, 20, 30]
W = 50
n = len(val)
print(knapSack(W, wt, val, 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[] val = new int[] { 60, 100, 120 };
        int[] wt = new int[] { 10, 20, 30 };
        int W = 50;
        int n = val.Length;
 
        Console.WriteLine(knapSack(W, wt, val, n));
    }
}
 
// This code is contributed by Sam007


PHP




<?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
    $val = array(60, 100, 120);
    $wt = array(10, 20, 30);
    $W = 50;
    $n = count($val);
    echo knapSack($W, $wt, $val, $n);
     
// This code is contributed by Sam007.
?>


Javascript




<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 val = [ 60, 100, 120 ];
    let wt = [ 10, 20, 30 ];
    let W = 50;
    let n = val.length;
    document.write(knapSack(W, wt, val, n));
</script>


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’.

Below is the implementation of the same approach but with optimized space complexity:

C++




// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// we can further improve the above Knapsack function's
// space complexity
int knapSack(int W, int wt[], int val[], int n)
{
    int i, w;
    int K[2][W + 1];
    // We know we are always using the current row or
    // the previous row of the array/vector . Thereby we can
    // improve it further by using a 2D array but with only
    // 2 rows i%2 will be giving the index inside the bounds
    // of 2d array K
 
    for (i = 0; i <= n; i++) {
        for (w = 0; w <= W; w++) {
            if (i == 0 || w == 0)
                K[i % 2][w] = 0;
            else if (wt[i - 1] <= w)
                K[i % 2][w] = max(
                    val[i - 1]
                        + K[(i - 1) % 2][w - wt[i - 1]],
                    K[(i - 1) % 2][w]);
            else
                K[i % 2][w] = K[(i - 1) % 2][w];
        }
    }
    return K[n % 2][W];
}
 
// Driver Code
int main()
{
    int val[] = { 60, 100, 120 };
    int wt[] = { 10, 20, 30 };
    int W = 50;
    int n = sizeof(val) / sizeof(val[0]);
 
    cout << knapSack(W, wt, val, n);
 
    return 0;
}
 
// This code was improved by Udit Singla


Java




// Java program for the above approach
 
import java.util.*;
class GFG {
 
    // we can further improve the above Knapsack function's
    // space complexity
    static int knapSack(int W, int wt[], int val[], int n)
    {
        int i, w;
        int[][] K = new int[2][W + 1];
 
        // We know we are always using the current row or
        // the previous row of the array/vector . Thereby we
        // can improve it further by using a 2D array but
        // with only 2 rows i%2 will be giving the index
        // inside the bounds of 2d array K
        for (i = 0; i <= n; i++) {
            for (w = 0; w <= W; w++) {
                if (i == 0 || w == 0)
                    K[i % 2][w] = 0;
                else if (wt[i - 1] <= w)
                    K[i % 2][w] = Math.max(
                        val[i - 1]
                            + K[(i - 1) % 2][w - wt[i - 1]],
                        K[(i - 1) % 2][w]);
                else
                    K[i % 2][w] = K[(i - 1) % 2][w];
            }
        }
        return K[n % 2][W];
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int val[] = { 60, 100, 120 };
        int wt[] = { 10, 20, 30 };
        int W = 50;
        int n = val.length;
 
        System.out.print(knapSack(W, wt, val, n));
    }
}
 
// This code is contributed by gauravrajput1


Python3




# we can further improve the above Knapsack function's space
# complexity
 
 
def knapSack(W, wt, val, n):
 
    K = [[0 for x in range(W+1)] for y in range(2)]
 
    # We know we are always using the  current row or
    # the previous row of the array/vector . Thereby we can
    # improve it further by using a 2D array but with only
    # 2 rows i%2 will be giving the index inside the bounds
    # of 2d array K
    for i in range(n + 1):
        for w in range(W + 1):
            if (i == 0 or w == 0):
                K[i % 2][w] = 0
            elif (wt[i - 1] <= w):
                K[i % 2][w] = max(
                    val[i - 1]
                    + K[(i - 1) % 2][w - wt[i - 1]],
                    K[(i - 1) % 2][w])
 
            else:
                K[i % 2][w] = K[(i - 1) % 2][w]
 
    return K[n % 2][W]
 
 
# Driver Code
if __name__ == "__main__":
 
    val = [60, 100, 120]
    wt = [10, 20, 30]
    W = 50
    n = len(val)
 
    print(knapSack(W, wt, val, n))
 
    # This code is contributed by ukasp.


C#




// C# program for the above approach
 
using System;
 
public class GFG {
 
    // we can further improve the above Knapsack function's
    // space complexity
    static int knapSack(int W, int[] wt, int[] val, int n)
    {
        int i, w;
        int[, ] K = new int[2, W + 1];
 
        // We know we are always using the  current row or
        // the previous row of the array/vector . Thereby we
        // can improve it further by using a 2D array but
        // with only 2 rows i%2 will be giving the index
        // inside the bounds of 2d array K
        for (i = 0; i <= n; i++) {
            for (w = 0; w <= W; w++) {
                if (i == 0 || w == 0)
                    K[i % 2, w] = 0;
                else if (wt[i - 1] <= w)
                    K[i % 2, w] = Math.Max(
                        val[i - 1]
                            + K[(i - 1) % 2, w - wt[i - 1]],
                        K[(i - 1) % 2, w]);
                else
                    K[i % 2, w] = K[(i - 1) % 2, w];
            }
        }
        return K[n % 2, W];
    }
 
    // Driver Code
    public static void Main(String[] args)
    {
        int[] val = { 60, 100, 120 };
        int[] wt = { 10, 20, 30 };
        int W = 50;
        int n = val.Length;
 
        Console.Write(knapSack(W, wt, val, n));
    }
}
 
// This code is contributed by umadevi9616


Javascript




<script>
    // we can further improve the above Knapsack function's space
    // complexity
    function knapSack(W , wt , val , n) {
        var i, w;
        var K = Array(2).fill().map(()=>Array(W + 1).fill(0));
 
        // We know we are always using the current row or
        // the previous row of the array/vector . Thereby we can
        // improve it further by using a 2D array but with only
        // 2 rows i%2 will be giving the index inside the bounds
        // of 2d array K
        for (i = 0; i <= n; i++) {
            for (w = 0; w <= W; w++) {
                if (i == 0 || w == 0)
                    K[i % 2][w] = 0;
                else if (wt[i - 1] <= w)
                    K[i % 2][w] = Math.max(val[i - 1] +
                    K[(i - 1) % 2][w - wt[i - 1]],
                    K[(i - 1) % 2][w]);
                else
                    K[i % 2][w] = K[(i - 1) % 2][w];
            }
        }
        return K[n % 2][W];
    }
 
    // Driver Code
        var val = [ 60, 100, 120 ];
        var wt = [ 10, 20, 30 ];
        var W = 50;
        var n = val.length;
 
        document.write(knapSack(W, wt, val, n));
 
// This code is contributed by Rajput-Ji
</script>


Output

220

Time Complexity: O(N * W)
Auxiliary Space: O(2 * W) As we are using a 2-D array but with only 2 rows

0-1 Knapsack Problem using dynamic programming(Space optimized):

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 <bits/stdc++.h>
using namespace std;
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]);
        }
    }
    return dp[W]; // returning the maximum value of knapsack
}
 
// Driver code
int main()
{
    int val[] = { 60, 100, 120 };
    int wt[] = { 10, 20, 30 };
    int W = 50;
    int n = sizeof(val) / sizeof(val[0]);
    cout << knapSack(W, wt, val, 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]);
            }
        }
        return dp[W]; // returning the maximum value of
                      // knapsack
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int val[] = { 60, 100, 120 };
        int wt[] = { 10, 20, 30 };
        int W = 50;
        int n = val.length;
        System.out.print(knapSack(W, wt, val, n));
    }
}
 
// This code is contributed by gauravrajput1


Python3




# code
# 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):
    dp = [0 for i in range(W+1)]  # Making the dp array
 
    for i in range(1, n+1):  # taking first i elements
        for w in range(W, 0, -1):  # starting from back,so that we also have data of
                                # previous computation when taking i-1 items
            if wt[i-1] <= w:
                # finding the maximum value
                dp[w] = max(dp[w], dp[w-wt[i-1]]+val[i-1])
 
    return dp[W]  # returning the maximum value of knapsack
 
 
# Driver code
val = [60, 100, 120]
wt = [10, 20, 30]
W = 50
n = len(val)
# This code is contributed by Suyash Saxena
print(knapSack(W, wt, val, n))


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]);
            }
        }
        return dp[W]; // returning the maximum value of
                      // knapsack
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        int[] val = { 60, 100, 120 };
        int[] wt = { 10, 20, 30 };
        int W = 50;
        int n = val.Length;
        Console.Write(knapSack(W, wt, val, n));
    }
}
 
// This code is contributed by gauravrajput1


Javascript




<script>
    function knapSack(W , wt , val , n)
    {
     
        // making and initializing dp array
        var dp = Array(W + 1).fill(0);
 
        for (i = 1; i < n + 1; i++) {
            for (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]);
            }
        }
        return dp[W]; // returning the maximum value of knapsack
    }
 
    // Driver code
        var val = [ 60, 100, 120 ];
        var wt = [ 10, 20, 30 ];
        var W = 50;
        var n = val.length;
        document.write(knapSack(W, wt, val, n));
 
// This code is contributed by Rajput-Ji
</script>


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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