Sequences:
We were discussing sequence databases.  Sequence databases support specialized processing of sequences with the help of new data structures that are usually not found in traditional storage systems. Sequences tend to number in millions if not more.  In this section, we focus on the storage concerns for sequences.

Sequence storage can enable different search operators to run against the same storage. If there is a narrow range of sequences that are to be evaluated, even a LogParser like utility can enable SQL commands to be run against the sequences.

It is important to note that
the language for querying of data remains the same whether they are sequences or entities. A lot of commercial analysis stacks are built on top of the standard query language. Even databstores that are not relational tend to pick up an adapter that facilitates SQL  This is more for the convenience of the developers rather than any other requirement. This query language is supported by standard query operators in most languages. Many developers are able to switch the data source without modifying their queries. This lets them write their applications once and run them anywhere.

#codingexercise
Sort integers of the array A1 by those of A2.

public List <Integer> sort (List <Integer> A1, List <Integer> A2)  { 

       List <Integer> result = new ArrayList <>(); 

       int start = 0; 

    

       A1.sort(); 

       for (Integer a: A2) { 

               for (Integer index = binarySearch(A1, start, A1.size()-1, a); start < A1.size () && index != -1; start = index +1) { 

                       result.add (A1 [index]); 

                } 

                start = 0; 

        } 

         if ( result.size () < A1.size () ) { 

                result.addAll (A1.getRange (result.size ()).sort ()); 

         } 

         return result; 

} 

int binarySearch(List <integer>  input, int start, int end, int val)
{
If (start > end) return –1;
int mid = (start + end)/2;
if (input[mid] == val) return mid;
if (start == end && input[mid] != val) return -1;
if (input[mid] < val)
return binarySearch(nums, mid+1, end, val);
else
return binarySearch(nums, start, mid, val);
}