Knapsack Problem 0 or 1

/**
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 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).
*/

public class KnapSack01{

 
   // 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) //if this item fits with available weight; wt[i-1] because weight index starts with zero
                   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];

}

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)); //220
    }
}