Java Anagram Buster
While browsing the net, I found this problem somewhere – to write a code that tests the given two strings are anagrams or not. From Wiki,
“An anagram is a type of word play, the result of rearranging the letter of a word or phrase to produce a new word or phrase, using all the original letters exactly once; for example orchestra can be rearranged into carthorse.”
I tried to solve this problem using Java and below is the result of it. The algorithm I tried is very simple:
- Clean the input – remove all the spaces and punctuation marks (because it doesn’t affect the compassion).
- Go through character by character from string one and check if that character exists in string two.
- If exists, then remove it from string two and move on to next character. If not exists, then we found a mismatch and the string is not an anagram.
- If all character from string one exists in string two, then we found it’s an anagram.
Java code to test two strings are anagrams:
public class AnagramTester {
public static void main(String[] args) {
String one = "The United States of America";
String two = "Attaineth its cause, freedom";
System.out.println(new AnagramTester().test(one, two));
}
public boolean test(String a, String b) {
boolean result = true;
StringBuilder one = new StringBuilder(a.replaceAll("[\\s+\\W+]", "").toLowerCase());
StringBuilder two = new StringBuilder(b.replaceAll("[\\s+\\W+]", "").toLowerCase());
if (one.length() == two.length()) {
int index = -1;
for (char c : one.toString().toCharArray()) {
index = two.indexOf(String.valueOf(c));
if (index == -1) {
result = false;
break;
}
two.deleteCharAt(index);
}
} else {
result = false;
}
return result;
}
}
I tested the above code with most of the anagrams found in the Anagram Site and it worked well.
If you think the above code can be improved in someway, feel free to comment.
(Note: Opinions expressed in this article and its replies are the opinions of their respective authors and not those of DZone, Inc.)
Comments
Carlos Hoces replied on Sun, 2012/03/18 - 5:27pm
Another possible solution:
public boolean test(String a, String b) {
boolean result = false;
final String one = a.replaceAll("[\\s+\\W+]", "").toLowerCase();
final String two = b.replaceAll("[\\s+\\W+]", "").toLowerCase();
if (one.length() == two.length()) {
final char[] oneArray = one.toCharArray();
final char[] twoArray = two.toCharArray();
Arrays.sort(oneArray);
Arrays.sort(twoArray);
result = Arrays.equals(oneArray, twoArray);
}
return result;
}
Francesco Illuminati replied on Sun, 2012/03/18 - 2:02pm
Considering the prime numbers calculation out of the algorithm cost this seems like an efficient alternative:
public class Prime { public static int[] calculateFirst(final int number) { final int[] primes = new int[number]; int counter = 1, index = 0; boolean isPrime; while (index < number) { counter++; isPrime = true; for (int i=0; i<index; i++) { if (counter % primes[i] == 0) { isPrime = false; break; } } if (isPrime) { primes[index] = counter; index++; } } return primes; } } public class AlphabetIndexer { public final static int ALPHABET_LENGHT = 'Z' - 'A' + 1; public static int getIndex(final char c) { int value = c - 'A'; if (value < 0 || value > ALPHABET_LENGHT) { value = c - 'a'; if (value < 0 || value > ALPHABET_LENGHT) { return -1; } } return value; } } public class Anagram { private final static int[] PRIMES = Prime.calculateFirst('Z' - 'A' + 1); public static int calculateValue(final String phrase) { int value = 1; for (char c: phrase.toCharArray()) { int index = AlphabetIndexer.getIndex(c); if (index != -1) { value += PRIMES[index]; } } return value; } public static boolean isAnagram(final String phrase1, final String phrase2) { return calculateValue(phrase1) == calculateValue(phrase2); } }David Whatever replied on Sun, 2012/03/18 - 4:43pm
Since you have to walk the strings anyway, I would say a good algorithm would be:
Jonas Olsson replied on Mon, 2012/03/19 - 5:17am
in response to:
David Whatever
Francesco Illuminati replied on Mon, 2012/03/19 - 5:19am
in response to:
David Whatever
David Whatever replied on Tue, 2012/03/20 - 12:59am
in response to:
Francesco Illuminati
For a more clever solution, you could change the methods slightly to expect the 26-int array to be passed in, and increment each index based on the count of each corresponding letter. Then the process changes to:
- Perform the easy checks
- Initialize 26-int array (defaults to zeroes)
- Pass first string and array into method to increment array based on letter frequency
- Negate each index in the array
- Pass second string and array into method to increment array based on letter frequency
- Verify array is zeroed out
I don't think the change is worth saving the creation of a second array, though.