AnagramDictionary.java

// Name: Aayush Khanna	
// USC NetID: aayushkh	
// CS 455 PA4
// Spring 2017

import java.io.FileNotFoundException;
import java.util.ArrayList;
import java.util.HashMap;
import java.util.Map;
import java.util.TreeMap;
import java.io.BufferedReader;
import java.io.FileReader;
import java.io.IOException;

/**
 * A dictionary of all anagram sets. Note: the processing is case-sensitive; so
 * if the dictionary has all lower case words, you will likely want any string
 * you test to have all lower case letters too, and likewise if the dictionary
 * words are all upper case.
 */

public class AnagramDictionary {

	private final static int ZERO = 0;
	private final static int ONE = 1;
	private Map<Character, Integer> singleWord;
	private Map<Map<Character, Integer>, ArrayList<String>> dictionaryMap;

	/**
	 * Create an anagram dictionary from the list of words given in the file
	 * indicated by fileName. PRE: The strings in the file are unique.
	 * 
	 * @param fileName
	 *            the name of the file to read from
	 * @throws FileNotFoundException
	 *             if the file is not found
	 */

	public AnagramDictionary() {
		singleWord = new TreeMap<Character, Integer>();
		dictionaryMap = new HashMap<Map<Character, Integer>, ArrayList<String>>();
	}

	/**
	 * Gets all anagrams of the given string. This method is case-sensitive.
	 * E.g. "CARE" and "race" would not be recognized as anagrams. Checks if
	 * tempWord is a key in the dictionaryMap. Hence the time complexity is
	 * O(1).
	 * 
	 * @param s
	 *            string to process
	 * @param tempWord
	 *            stores the characters of String s as key and number of
	 *            instances of the character as value in the Map.
	 * @return a list of the anagrams of s
	 * 
	 */
	
	public AnagramDictionary(String fileName) throws IOException {
		try (BufferedReader br = new BufferedReader(new FileReader(fileName))) {
			String line;
			ArrayList<String> temp = new ArrayList<String>();
			dictionaryMap = new HashMap<Map<Character, Integer>, ArrayList<String>>();
			while ((line = br.readLine()) != null) {
				singleWord = new TreeMap<Character, Integer>();
				for (int i = 0; i < line.length(); i++) {
					char letter = line.charAt(i);
					int count = singleWord.containsKey(letter) ? singleWord.get(letter) : ZERO;
					singleWord.put(letter, count + ONE);
				}
				temp = dictionaryMap.containsKey(singleWord) ? dictionaryMap.get(singleWord) : new ArrayList<String>();
				temp.add(line);
				dictionaryMap.put(singleWord, temp);
			}
		}
	}

	/**
	 * Gets all anagrams of the given string. This method is case-sensitive.
	 * E.g. "CARE" and "race" would not be recognized as anagrams. Checks if
	 * tempWord is a key in the dictionaryMap. Hence the time complexity is
	 * O(1).
	 * 
	 * @param s
	 *            string to process
	 * @param tempWord
	 *            stores the characters of String s as key and number of
	 *            instances of the character as value in the Map.
	 * @return a list of the anagrams of s
	 * 
	 */
	public ArrayList<String> getAnagramsOf(String s) {
		String convertedString = s.toLowerCase();
		Map<Character, Integer> tempWord = new TreeMap<Character, Integer>();
		for (int i = 0; i < convertedString.length(); i++) {
			char letter = convertedString.charAt(i);
			int count = tempWord.containsKey(letter) ? tempWord.get(letter) : ZERO;
			tempWord.put(letter, count + ONE);
		}
		ArrayList<String> temp = dictionaryMap.containsKey(tempWord) ? dictionaryMap.get(tempWord)
				: new ArrayList<String>();
		return temp;
	}

}