Java Program 0-1 Knapsack Problem

Recursive Solution

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/* 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 program to test above function
    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 */

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Output:

220

Dynamic Programming Solution

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// A Dynamic Programming based solution for 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)
    {
        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 program to test above function
    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 */

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Output:

220

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



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