Today we continue discussing the best practice from storage engineering :

466) Storage products have multiple paths of data entry usually via protocols. These are each tested using their respective protocol tools

467) Storage products are usually part of tiered storage. As such data aging and validation needs to be covered

468) Storage products are tested with different batches of loads. They are also tested using continuous loads with varying rate over time

469) Storage products are tested for size and usage under all circumstances. These can be fine grained or aggregated and can be queried at different scopes and levels.

470) Storage product is generally the source of truth for all upstream data sources and workflows.

#codingexercise
The coin selection problem can scale to any constant number of coins that can be picked in each turn using the methods below.
int GetCoinsKWithDP(List<int> coins, int i, int j, int k)
{
if (i > j) return 0;
if (i==j) return coins[i];
if (j - i +1 <= k) {
     List <int>  change = new ArrayList <int>();
    for (int c = i ; c <= j; c++) {
           change.add (coins [c]) ;
    }
    return Collections.sum (change);
}
List <int> options = new ArrayList ();
for (int m = 0; m < k; m++) {
      int change = 0;

      for (int left = 0; left < k; left++) {
             change += coins [i+left];   
             int option = change + GetCoinsDP (sequence, i+left, j);
     options.add (option);
      }

      for (int right = 0; right < k; right++) {
             change += coins [j-right];
int option = change + GetCoinsDP (sequence, i, j-right);
     options.add (option);
      }
   
        for (int left = 0; left < k; left++) {
             for (int right = 0; right < k-left right++) {
             change += coins [j-right];
int option = change + GetCoinsDP (sequence, i+left, j-right);
     options.add (option);
      }
}
}
return Collections.max (options);
}
The selections in each initiation level of this GetCoinsKWithDP can be added to a list and alternate additions can be skipped as belonging to the other player since the method remains the same for both.