content: 
  // 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));
      }
  }