Writing an anagram finder (from a word list in a txt file)

Question

I've been trying to code an anagram finder in Java so that in the terminal after compiling all I would have to do is

1. Type in java Anagramfind list.txt
2. When prompted type in a word, say treasure
3. The program prints an anagram, like austerer
4. Another prompt comes up asking if I would like another one (yes/no)

The list.txt file has most, if not all, of the words in the English language.

Here's what I have so far...

import java.util.*;
import java.io.*;

class ProjectAnagram {
    public static void main (String[] args throws IOException) {
        //THis here declares an array of strings
        Scanner dictionary = new Scanner new (fileInputStream(args[0]));
        String[] entireArray = new String[173528]; //name of array + 173258
        System.out.println ("Put something in please");
        Scanner keyboard = new Scanner(System.in);
        System.out.println ("Inserted");

        String word = keyboard;
    }

I still need to add the rest.

I've mostly been having trouble with the use of arrays, which I referenced here:

• Java read file and store text in an array
• http://docs.oracle.com/javase/tutorial/essential/environment/cmdLineArgs.html
• What is "String args[]"? parameter in main method Java

I also am having trouble using a Stringbuffer which to check if words have the same characters or not.

The program checks if the input string and if the string in the text file have the same length first so as to rule out obvious non-anagrams. If it doesn't, then it moves on to the next word in the list, probably with i++ in some loop.

Answer 1

Similar to finding if two strings are permutations of each other, you can sort the characters of the given string and make it the key to a list of anagrams. This way no matter the string you will find only strings of the same length comprised of the same characters.

Something like:

Map<String, List<String>> map ...
map.get(getKey(string)).get(i); // i = the ith request for an anagram

Answer 2

Arrays.equals('Testing'.chars().sorted().toArray(), 'ingsetT'.chars().sorted().toArray())

Answer 3

Try this.

package stackoverflow;

import java.io.*;
import java.util.*;

public class ProjectAnagram {

    static String sort(String s) {
        char[] c = s.toCharArray();
        Arrays.sort(c);
        return String.valueOf(c);
    }

    public static void main(String[] args) throws FileNotFoundException {
        Map<String, List<String>> words = new HashMap<>();
        try (Scanner in = new Scanner(new File(args[0]))) {
            while (in.hasNext()) {
                String word = in.next();
                String sorted = sort(word);
                List<String> list = words.get(sorted);
                if (list == null)
                    words.put(sorted, list = new ArrayList<>());
                list.add(word);
            }
        }
        Scanner in = new Scanner(System.in);
        while (true) {
            System.out.print("Enter word (or press ENTER to quit): ");
            if (!in.hasNextLine()) break;
            String s = in.nextLine();
            if (s.length() == 0) break;
            System.out.println(words.get(sort(s)));
        }
    }
}

Answer 4

EDIT: This answer is an efficient way to do it!

Hashmaps are O(1) look up time.

We would need 3 iterations. First to add character count in first string to hashmap, second to remove character count in second string from hashmap, third to iterate through hashmap and see if all the values were 0.

So, this would give us an O(n) algorithm.