How to find the number of strings that are in the particular range FAST?

Question

I was given a question:

Given a list of strings, and a list of queries: START and END, both of which are strings. I have to find the number of strings that are in the range of [START, END)

For example: a list of strings: A, AA, AB, CD, ZS, XYZ a list of queries:

A, AA
A, CC
AB, ZZ
AC, CD

The output should be:

1
3
3
0

The way I approach this problem is that: while iterating through the list of strings, I create an AVL tree by inserting new string one by one. (At first, I used unbalanced BST but I got Time Limit.) When doing the comparison, I use compareTo function in java String.

After creating the AVL tree, I run the query that counts from [start, end). My method is that

1. let v = root.

2. if v==null -> return 0 

   else if v.value < start -> count(v.right)

   else if v.value >= end -> count(v.left)

   else 1 + count(v.right) + count(v.left)

However, I still got time limit pernalty :(

Therefore, I change method by creating hash function by hashing into double and instead of using compareTo, I compared the hash value instead.

But, I still got time limit!

So, I store the value of subtree size into each vertex, and instead of using count or the time, I add more conditional statements, some of which can use the size of the subtree instead of calling count function recursively.

Any suggestion to me to get it run in a particular time? :\\

Answer 1

Use an order statistic tree: http://en.wikipedia.org/wiki/Order_statistic_tree

It is basically a modified balanced bst where each node stores the size of its subtree, which allows you to answer queries about how many items there are before a given item in log(n) time.