CareerCup String getSentence(String text, Set<String> dictionary); O(n)

String getSentence(String text, Set<String> dictionary); 

// text is a string without spaces, you need to insert spaces into text, so each word seperated by the space in the resulting string exists in the dictionary, return the resulting string 

// running time has to be at least as good as O(n) 

 

 

// getSentence("iamastudentfromwaterloo", {"from, "waterloo", "hi", "am", "yes", "i", "a", "student"}) -> "i am a student from waterloo"

 

 

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import java.util.ArrayList;
import java.util.HashSet;
import java.util.List;
import java.util.Set;

class Solution {
	
	public static void main(String[] args) {
		
		Solution instance = new Solution();
		Set<String> dictionary = new HashSet();
		dictionary.add("google");
		dictionary.add("is");
		dictionary.add("awesome");
		List<String> store = new ArrayList<String>();
		instance.printWords("googleisawesome", store, dictionary);
		for(int i = store.size() - 1; i >= 0; --i) {
			System.out.println(store.get(i));
		}
	}


	private boolean printWords(String string, List<String> store, Set<String> dictionary) {
		if(string.length() == 0) {
			return true;
		}
		for(int i = 1; i <= string.length(); ++i) {
			String curWord = string.substring(0, i);
			if(dictionary.contains(curWord) && printWords(string.substring(i), store, dictionary)) {
				store.add(curWord);
				return true;
			}
		}
		return false;
	}
}

Trie tree is expected for this problem...