Comparing two words

Question

I am new to the forum ( as a registered user ), so I tried really hard ( I swear! ) not to post a question, and look for old answers, and I checked other people's mistakes in problems similar to mine but I couldn't fine anything.

My code, downhere, is supposed to check if a word is an anagram of another or not. I am pretty sure I have complicated my life a lot and there would have been easier ways to do it but... I have been working on this for some time now and would like to see it work...

Any idea why it doesn't?

All I see is empty dictionaries and the two words are always anagrams when they have the same number of letters ( meaning that in fact my dictionaries aren't doing anything :'( )

import acm.program.ConsoleProgram;
import java.util.*;


public class Anagrams extends ConsoleProgram {

    String firstWord;
    String secondWord;

    public boolean checkLength(String firstWord, String secondWord) {
        if (firstWord.length() == secondWord.length()) {
            println("Same length!");
            return true;
        } else {
            return false;
        }
    }

    public boolean anagram(String firstWord, String secondWord) {
        firstWord = firstWord.toLowerCase();
        secondWord = secondWord.toLowerCase();
        String[] firstArray = firstWord.split("\\a");
        String[] secondArray = secondWord.split("\\a");
        int firstLength = firstWord.length();
        int secondLength = secondWord.length();

        Map<String, Integer> firstDictionary = new HashMap<String, Integer>();
        Map<String, Integer> secondDictionary = new HashMap<String, Integer>();

        for (firstLength = 0; firstLength == firstArray.length; firstLength++) {
            System.out.println("checking the letter " + firstArray[firstLength] + " in array" + firstArray.toString());
            if (firstDictionary.get(firstArray[firstLength]) == null) {
                firstDictionary.put(firstArray[firstLength], 1);
            } else {
                firstDictionary.put(firstArray[firstLength], firstDictionary.get(firstArray[firstLength]) + 1);
            }
        }

        for (secondLength = 0; secondLength == secondArray.length; secondLength++) {
            if (secondDictionary.get(secondArray[secondLength]) == null) {
                secondDictionary.put(secondArray[secondLength], 1);
            } else {
                secondDictionary.put(secondArray[secondLength], secondDictionary.get(secondArray[secondLength]) + 1);
            }
        }

        if (firstDictionary.equals(secondDictionary)) {
            return true;
        } else {
            return false;
        }
    }

    public void run() {
        int runAgain = 0;
        while (runAgain == 0) {
            println("Enter the first word to be analyzed");
            firstWord = readLine();
            println("Enter the second word to be analyzed");
            secondWord = readLine();

            if (checkLength(firstWord, secondWord) == true) {

                if (anagram(firstWord, secondWord) == true) {
                    println("Yes! The two words are anagrams!");
                }
            } else {
                println("No. The two words are not anagrams!");
            }
        }

    }
}

Pham Trung · Accepted Answer · 2014-09-18T03:41:33.053

You have an error in your loop condition

  for (firstLength=0;firstLength==firstArray.length;firstLength++)

This should be

   for (firstLength=0;firstLength<firstArray.length;firstLength++)

Because the condition firstLength==firstArray.length is false, so your program will never enter the loop.

And I believe:

String[] firstArray = firstWord.split("\\a");
String[] secondArray = secondWord.split("\\a");

is wrong, you just need char array, so using this instead:

 char[] first = firstWord.toCharArray();
 char[] second = secondWord.toCharArray();

Working version:

 public boolean anagram(String firstWord, String secondWord) {
        firstWord = firstWord.toLowerCase();
        secondWord = secondWord.toLowerCase();
        char[] firstArray = firstWord.toCharArray();
        char[] secondArray = secondWord.toCharArray();
        int firstLength = firstWord.length();
        int secondLength = secondWord.length();

        Map<Character, Integer> firstDictionary = new HashMap<>();
        Map<Character, Integer> secondDictionary = new HashMap<>();

        for (firstLength = 0; firstLength < firstArray.length; firstLength++) {
           // System.out.println("checking the letter " + firstArray[firstLength] + " in array" + firstArray.toString());
            if (!firstDictionary.containsKey(firstArray[firstLength])) {
                firstDictionary.put(firstArray[firstLength], 1);
            } else {
                firstDictionary.put(firstArray[firstLength], firstDictionary.get(firstArray[firstLength]) + 1);
            }
        }

        for (secondLength = 0; secondLength < secondArray.length; secondLength++) {
            if (!secondDictionary.containsKey(secondArray[secondLength])) {
                secondDictionary.put(secondArray[secondLength], 1);
            } else {
                secondDictionary.put(secondArray[secondLength], secondDictionary.get(secondArray[secondLength]) + 1);
            }
        }

        if (firstDictionary.equals(secondDictionary)) {
            return true;
        } else {
            return false;
        }
    }

and everythig works after? – Kick Buttowski Sep 18 '14 at 03:39 — Kick Buttowski, Sep 18 '14 at 03:39
@KickButtowski yes, at least with some simple tests – Pham Trung Sep 18 '14 at 03:47 — Pham Trung, Sep 18 '14 at 03:47

score 0 · Answer 2 · answered Sep 18 '14 at 03:41

0

I would go with a different approach:

String firstWord = "abcde";
String secondWord = "edcba";
String[] arr1 = firstWord.split("(?!^)"); // split each string to an array of (String) characters
Arrays.sort(arr1); // and sort it alphabetically 
String[] arr2 = secondWord.split("(?!^)");
Arrays.sort(arr2);
// now we can compare:
System.out.println(Arrays.equals(arr1, arr2)); // prints 'true'

answered Sep 18 '14 at 03:41

Nir Alfasi

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That`s actually an interesting way of solving it. I did not think about that! – Francesco Brouge Nutricato Sep 18 '14 at 05:51

Comparing two words

2 Answers2