How to check if two words are anagrams

Question

I have a program that shows you whether two words are anagrams of one another. There are a few examples that will not work properly and I would appreciate any help, although if it were not advanced that would be great, as I am a 1st year programmer. "schoolmaster" and "theclassroom" are anagrams of one another, however when I change "theclassroom" to "theclafsroom" it still says they are anagrams, what am I doing wrong?

import java.util.ArrayList;
public class AnagramCheck
{
  public static void main(String args[])
  {
      String phrase1 = "tbeclassroom";
      phrase1 = (phrase1.toLowerCase()).trim();
      char[] phrase1Arr = phrase1.toCharArray();

      String phrase2 = "schoolmaster";
      phrase2 = (phrase2.toLowerCase()).trim();
      ArrayList<Character> phrase2ArrList = convertStringToArraylist(phrase2);

      if (phrase1.length() != phrase2.length()) 
      {
          System.out.print("There is no anagram present.");
      } 
      else 
      {
          boolean isFound = true;
          for (int i=0; i<phrase1Arr.length; i++)
          {  
              for(int j = 0; j < phrase2ArrList.size(); j++) 
              {
                  if(phrase1Arr[i] == phrase2ArrList.get(j))
                  {
                      System.out.print("There is a common element.\n");
                      isFound = ;
                      phrase2ArrList.remove(j);
                  }
              }
              if(isFound == false)
              {
                  System.out.print("There are no anagrams present.");
                  return;
              } 
          }
          System.out.printf("%s is an anagram of %s", phrase1, phrase2);
      }
  }

  public static ArrayList<Character> convertStringToArraylist(String str) {
      ArrayList<Character> charList = new ArrayList<Character>(); 
      for(int i = 0; i<str.length();i++){
          charList.add(str.charAt(i));
      }
      return charList;
  }
}

Answer 1

Here's a simple fast O(n) solution without using sorting or multiple loops or hash maps. We increment the count of each character in the first array and decrement the count of each character in the second array. If the resulting counts array is full of zeros, the strings are anagrams. Can be expanded to include other characters by increasing the size of the counts array.

class AnagramsFaster{

    private static boolean compare(String a, String b){
        char[] aArr = a.toLowerCase().toCharArray(), bArr = b.toLowerCase().toCharArray();
        if (aArr.length != bArr.length)
            return false;
        int[] counts = new int[26]; // An array to hold the number of occurrences of each character
        for (int i = 0; i < aArr.length; i++){
            counts[aArr[i]-97]++;  // Increment the count of the character at i
            counts[bArr[i]-97]--;  // Decrement the count of the character at i
        }
        // If the strings are anagrams, the counts array will be full of zeros
        for (int i = 0; i<26; i++)
            if (counts[i] != 0)
                return false;
        return true;
    }

    public static void main(String[] args){
        System.out.println(compare(args[0], args[1]));
    }
}

Answer 2

import java.util.ArrayList;
import java.util.List;
import java.util.Map;
import java.util.TreeMap;
/**
 * Check if Anagram by Prime Number Logic
 * @author Pallav
 *
 */
public class Anagram {
    public static void main(String args[]) {
        System.out.println(isAnagram(args[0].toUpperCase(),
                args[1].toUpperCase()));
    }
/**
 * 
 * @param word : The String 1
 * @param anagram_word : The String 2 with which Anagram to be verified
 * @return true or false based on Anagram
 */
    public static Boolean isAnagram(String word, String anagram_word) {
        //If length is different return false
        if (word.length() != anagram_word.length()) {
            return false;
        }
        char[] words_char = word.toCharArray();//Get the Char Array of First String
        char[] anagram_word_char = anagram_word.toCharArray();//Get the Char Array of Second String
        int words_char_num = 1;//Initialize Multiplication Factor to 1
        int anagram_word_num = 1;//Initialize Multiplication Factor to 1 for String 2
        Map<Character, Integer> wordPrimeMap = wordPrimeMap();//Get the Prime numbers Mapped to each alphabets in English
        for (int i = 0; i < words_char.length; i++) {
            words_char_num *= wordPrimeMap.get(words_char[i]);//get Multiplication value for String 1
        }
        for (int i = 0; i < anagram_word_char.length; i++) {
            anagram_word_num *= wordPrimeMap.get(anagram_word_char[i]);//get Multiplication value for String 2
        }

        return anagram_word_num == words_char_num;
    }
/**
 * Get the Prime numbers Mapped to each alphabets in English
 * @return
 */
    public static Map<Character, Integer> wordPrimeMap() {
        List<Integer> primes = primes(26);
        int k = 65;
        Map<Character, Integer> map = new TreeMap<Character, Integer>();
        for (int i = 0; i < primes.size(); i++) {
            Character character = (char) k;
            map.put(character, primes.get(i));
            k++;
        }
        // System.out.println(map);
        return map;
    }
/**
 * get first N prime Numbers where Number is greater than 2
 * @param N : Number of Prime Numbers
 * @return
 */
    public static List<Integer> primes(Integer N) {
        List<Integer> primes = new ArrayList<Integer>();
        primes.add(2);
        primes.add(3);

        int n = 5;
        int k = 0;
        do {
            boolean is_prime = true;
            for (int i = 2; i <= Math.sqrt(n); i++) {
                if (n % i == 0) {
                    is_prime = false;
                    break;
                }
            }

            if (is_prime == true) {
                primes.add(n);

            }
            n++;
            // System.out.println(k);
        } while (primes.size() < N);

        // }

        return primes;
    }

}

Answer 3

We're walking two equal length strings and tracking the differences between them. We don't care what the differences are, we just want to know if they have the same characters or not. We can do this in O(n/2) without any post processing (or a lot of primes).

public class TestAnagram {
  public static boolean isAnagram(String first, String second) {
    String positive = first.toLowerCase();
    String negative = second.toLowerCase();

    if (positive.length() != negative.length()) {
      return false;
    }

    int[] counts = new int[26];

    int diff = 0;

    for (int i = 0; i < positive.length(); i++) {
      int pos = (int) positive.charAt(i) - 97; // convert the char into an array index
      if (counts[pos] >= 0) { // the other string doesn't have this
        diff++; // an increase in differences
      } else { // it does have it
        diff--; // a decrease in differences
      }
      counts[pos]++; // track it

      int neg = (int) negative.charAt(i) - 97;
      if (counts[neg] <= 0) { // the other string doesn't have this
        diff++; // an increase in differences
      } else { // it does have it
        diff--; // a decrease in differences
      }
      counts[neg]--; // track it
    }

    return diff == 0;
  }

  public static void main(String[] args) {
    System.out.println(isAnagram("zMarry", "zArmry")); // true
    System.out.println(isAnagram("basiparachromatin", "marsipobranchiata")); // true
    System.out.println(isAnagram("hydroxydeoxycorticosterones", "hydroxydesoxycorticosterone")); // true
    System.out.println(isAnagram("hydroxydeoxycorticosterones", "hydroxydesoxycorticosterons")); // false
    System.out.println(isAnagram("zArmcy", "zArmry")); // false
  }
}

Yes this code is dependent on the ASCII English character set of lowercase characters but it shouldn't be hard to modify to other languages. You can always use a Map[Character, Int] to track the same information, it'll just be slower.

Answer 4

A way to solve this - based on Sai Kiran's answer..

import java.util.Scanner;

public class Anagram {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);

        System.out.print("Enter first word : ");
        String word1 = sc.nextLine();
        System.out.print("Enter second word : ");
        String word2 = sc.nextLine();

        sc.close();

        System.out.println("Is Anagram : " + isAnagram(word1, word2));
    }

    private static boolean isAnagram(String word1, String word2) {

        if (word1.length() != word2.length()) {
            System.err.println("Words length didn't match!");
            return false;
        }

        char ch1, ch2;
        int len = word1.length(), sumOfWord1Chars = 0, sumOfWord2Chars = 0;

        for (int i = 0; i < len; i++) {
            ch1 = word1.charAt(i);
            if (word2.indexOf(ch1) < 0) {
                System.err.println("'" + ch1 + "' not found in \"" + word2
                        + "\"");
                return false;
            }
            sumOfWord1Chars += word1.charAt(i);

            ch2 = word2.charAt(i);
            if (word1.indexOf(ch2) < 0) {
                System.err.println("'" + ch2 + "' not found in \"" + word1
                        + "\"");
                return false;
            }
            sumOfWord2Chars += word2.charAt(i);
        }

        if (sumOfWord1Chars != sumOfWord2Chars) {
            System.err
                    .println("Sum of both words didn't match, i.e., words having same characters but with different counts!");
            return false;
        }

        return true;
    }
}

Answer 5

IMHO, the most efficient solution was provided by @Siguza, I have extended it to cover strings with space e.g: "William Shakespeare", "I am a weakish speller", "School master", "The classroom"

public int getAnagramScore(String word, String anagram) {

        if (word == null || anagram == null) {
            throw new NullPointerException("Both, word and anagram, must be non-null");
        }

        char[] wordArray = word.trim().toLowerCase().toCharArray();
        char[] anagramArray = anagram.trim().toLowerCase().toCharArray();

        int[] alphabetCountArray = new int[26];

        int reference = 'a';

        for (int i = 0; i < wordArray.length; i++) {
            if (!Character.isWhitespace(wordArray[i])) {
                alphabetCountArray[wordArray[i] - reference]++;
            }
        }
        for (int i = 0; i < anagramArray.length; i++) {
            if (!Character.isWhitespace(anagramArray[i])) {
                alphabetCountArray[anagramArray[i] - reference]--;
            }
        }

        for (int i = 0; i < 26; i++)
            if (alphabetCountArray[i] != 0)
                return 0;

        return word.length();

    }

Answer 6

Many complicated answers here. Base on the accepted answer and the comment mentioning the 'ac'-'bb' issue assuming A=1 B=2 C=3, we could simply use the square of each integer that represent a char and solve the problem:

public boolean anagram(String s, String t) {
    if(s.length() != t.length())
        return false;

    int value = 0;
    for(int i = 0; i < s.length(); i++){
        value += ((int)s.charAt(i))^2;
        value -= ((int)t.charAt(i))^2;
    }
    return value == 0;
}