package DynamicProgramming;


import java.util.Arrays;

/**
 * Recursive Solution for 0-1 knapsack with memoization
 */
public class KnapsackMemoization {

  private static int[][] t;


  // Returns the maximum value that can
  // be put in a knapsack of capacity W
  public static int knapsack(int[] wt, int[] value, int W, int n) {
    if(t[n][W] != -1) {
      return t[n][W];
    }
    if (n == 0 || W == 0) {
      return 0;
    }
    if (wt[n - 1] <= W) {
      t[n-1][W-wt[n-1]] = knapsack(wt, value, W - wt[n - 1], n - 1);
      // Include item in the bag. In that case add the value of the item and call for the remaining items
      int tmp1 = value[n - 1] + t[n-1][W-wt[n-1]];
      // Don't include the nth item in the bag anl call for remaining item without reducing the weight
      int tmp2 = knapsack(wt, value, W, n - 1);
      t[n-1][W] = tmp2;
      // include the larger one
      int tmp = tmp1 > tmp2 ? tmp1 : tmp2;
      t[n][W] = tmp;
      return tmp;
      // If Weight for the item is more than the desired weight then don't include it
      // Call for rest of the n-1 items
    } else if (wt[n - 1] > W) {
      t[n][W] = knapsack(wt, value, W, n - 1);
      return t[n][W];
    }
    return -1;
  }

  // Driver code
  public static void main(String args[]) {
    int[] wt = {1, 3, 4, 5};
    int[] value = {1, 4, 5, 7};
    int W = 10;
    t = new int[wt.length+1][W+1];
    Arrays.stream(t).forEach(a -> Arrays.fill(a, -1));
    int res = knapsack(wt, value, W, wt.length);
    System.out.println("Maximum knapsack value " + res);
  }
}