Java Program 0-1 Knapsack Problem

Last Updated : 09 Nov, 2023

Write a Java program for a 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.

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

Java Program for 0-1 Knapsack Problem using Recursion:

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.

Step-by-step approach:

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:

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 */
 
  

Output

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

Java Program for 0-1 Knapsack Problem using Dynamic Programming:

Memoization Approach for 0/1 Knapsack 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:

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
 
  

Output

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:

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.

Below is the implementation of the above approach:

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 */
 
  

Output

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:

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:

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
 
  

Output

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

Please refer complete article on Dynamic Programming | Set 10 ( 0-1 Knapsack Problem) for more details!