# Vmware Interview Question

This is the question recently asked for the Vmware MTS-2 Position from me.

You will be given a list of articles with their page lengths and intellectual value coefficients. Given a limit to the number of pages you can read in a day, determine the maximum intellectual value you can achieve during one day.
For example, your articles are of lengths articles =[2,2,3,4] and they are of intellectual value iv =[2,4,4,5]. If you can read p=15 pages in a day, what should you read? You have to read each article twice to gain value.Associating 2*articles[i] with iv[i], The maximum combined length of articles read twice is14 pages and there are two ways to get there: readarticles, articles and articles for a total intellectual value of 2 + 4 + 4 = 10 or read articlesand articles for a total of 4 + 5 = 9. Our maximal learning is 10 intellectual value points.

Complete the function maximum learning. The function must return the integer value representing the maximum intellectual value you can get in one day of reading.

This is the simple question of 0–1 Knapsack.

Below is the implementation.

`public static int maximumLearning(List<Integer> iv, List<Integer> articles, int p) {  int size = articles.size();  int[] art = new int[size];  int[] ivs = new int[size];for(int i=0;i<size;i++){   art[i] = articles.get(i)*2;   ivs[i] = iv.get(i);}return knapSackTopDownCode(ivs, art, p, size);}public static int knapSackTopDownCode(int[] val, int[] wt, int W, int n) {   int mat[][] = new int [n+1][W+1];   for(int i=1;i<n+1;i++) {     for(int j=1;j<W+1;j++) {          if(i==0 || j ==0)              mat[i][j] =0;          else if(wt[i-1]<=j) {              mat[i][j] = Math.max(val[i-1]+ mat[i-1][j-wt[i-       1]],mat[i-1][j]);           }          else {              mat[i][j] = mat[i-1][j];}             }       }   return mat[n][W]; }}`

Happy Coding!!

## More from Vipul Gupta

Software Developer