Fast substring in scala

Question

According to Time complexity of Java's substring() , java's substring takes linear time. Is there a faster way (may be in some cases)? I may suggest iterator, but suspect that it also takes O(n).

val s1: String  = s.iterator.drop(5).mkString

But several operations on an iterator would be faster than same operations on string, right?

Answer 1

If you need to edit very long string, consider using data structure called Rope .

Scalaz library has Cord class which is implementation of modified version of Rope.

A Cord is a purely functional data structure for efficiently storing and manipulating String s that are potentially very long. Very similar to Rope[Char] , but with better constant factors and a simpler interface since it's specialized for String s.

Answer 2

As Strings are - according to the linked question - always backed by a unique character array, substring can't be faster than O(n). You need to copy the character data.

As for alternatives: there will at least be one operation which is O(n). In your example, that's mkString which collects the characters in the iterator and builds a string from them.

However, I wouldn't worry about that too much. The fact that you're using a high level language means (should mean) that developer time is more valuable than CPU time for your particular task. substring is also the canonical way to ... take a substring, so using it makes your program more readable.

EDIT: I also like this sentence (from this answer ) a lot: O(n) is O(1) if n does not grow large. What I take away from this is: you shouldn't write inefficient code, but asymptotical efficiency is not the same as real-world efficiency.