// Ray McDougall
// 11/22/13
// CSE 143x
// TA: Kevin Quinn
// Programming Assignment #9
//
// This management class for the AnagramMain program will search a
// given word list for all possible sets of anagrams of a given phrase,
// up to a given word count cap. As this is performed through an
// exhaustive search, all possible permutations of each anagram combination
// are returned.

import java.util.*;

public class AnagramSolver {
	List<String> dictionary;
	Map<String, LetterInventory> inventories;
	
	// initializes the necessary word lists; stores each word
	// in the given dictionary with its letter inventory
	public AnagramSolver(List<String> list) {
		dictionary = new ArrayList<String>();
		inventories = new HashMap<String, LetterInventory>();
		for (String word : list) {
			dictionary.add(word);
			LetterInventory letters = new LetterInventory(word);
			inventories.put(word, letters);
		}
	}
	
	// pre: the max word count must not be negative, otherwise throws IllegalArgumentException
	// post: returns all anagram combinations up to the given
	// max word count
	public void print(String s, int max) {
		if (max < 0) {
			throw new IllegalArgumentException();
		}
		int level;
		// ensure that a max of zero will be a special case
		if (max == 0) {
			level = -1;
		} else {
			level = max;
		}
		LetterInventory target = new LetterInventory(s);
		prune(dictionary, target);
		List<String> result = new ArrayList<String>();
		search(target, result, level + 1);
	}
	
	// clears the word list of any unnecessary words (no possible anagrams)
	private static void prune(List<String> list, LetterInventory target) {
		for (String word : list) {
			LetterInventory temp = new LetterInventory(word);
			if (target.subtract(temp) == null) {
				list.remove(temp);
			}
		}
	}
	
	// prints the output of each possible anagram set directly to the console
	private void search(LetterInventory target, List<String> result, int level) {
		// make a pass through each word in the pruned dictionary
		for (Map.Entry<String, LetterInventory> currentEntry : inventories.entrySet()) {
			// make sure the current word will fit into the possible anagram set
			// and we are not going too many levels down (based on the max word count)
			if (target.subtract(currentEntry.getValue()) != null && level != 1) {
				result.add(currentEntry.getKey());
				// outputs a solution if it has been found, otherwise go down another level
				if (target.subtract(currentEntry.getValue()).isEmpty()) {
					System.out.println(result.toString());
				} else {
					search(target.subtract(currentEntry.getValue()), result, level - 1);
				}
				result.remove(currentEntry.getKey());
			}
		}
	}
}