During a recent job interview, I was asked to give a solution to the following problem:
Given a string s (without spaces) and a dictionary, return the words in the dictionary that compose the string.
For example, s= peachpie, dic= {peach, pie}, result={peach, pie}.
I will ask the the decision variation of this problem:
if s can be composed of words in the
dictionary return yes, otherwise
return no.
My solution to this was in backtracking (written in Java)
public static boolean words(String s, Set<String> dictionary)
{
if ("".equals(s))
return true;
for (int i=0; i <= s.length(); i++)
{
String pre = prefix(s,i); // returns s[0..i-1]
String suf = suffix(s,i); // returns s[i..s.len]
if (dictionary.contains(pre) && words(suf, dictionary))
return true;
}
return false;
}
public static void main(String[] args) {
Set<String> dic = new HashSet<String>();
dic.add("peach");
dic.add("pie");
dic.add("1");
System.out.println(words("peachpie1", dic)); // true
System.out.println(words("peachpie2", dic)); // false
}
What is the time complexity of this solution?
I'm calling recursively in the for loop, but only for the prefix's that are in the dictionary.
Any idea's?
