Write a Java Program to Test Given Words are Anagram or Not

Description

Two words are said to be anagrams if both the words contain the same set of characters with all original letters exactly once. For example, the word program can be re-arranged as grampor and these both words form an anagram. Following is a java program to check if a string is an anagram or not.

We will discuss 3 different ways to find if given words are anagram or not - one is by sorting and applying equals() method provided in Arrays class, another by comparing each character of the words and the other by filling the character count within an array of length 26(assuming the word may contain 26 small letters).

Method 1: Using Arrays.equals Method

This is a very simple approach. At first let us sort both the words. Java internally uses Quick sort for sorting an Array with an average time complexity of nlog(n) where n is the number of characters in the array.

Next, we can check if both the arrays are equal. If both the arrays are equal means the given words are anagrams string.

public class StringAnagram {

	public static void main(String[] args) {
		String input1= "cat";
		String input2 = "act";
		isAnagram(input1, input2);

	}
	
	public static void isAnagram(String input1, String input2){
        //Remove all whitespace first
        String  s1= input1.replaceAll("\s", "");
 
        String s2 = input2.replaceAll("\s", "");
        
        boolean status = true;
 
        if(s1.length() != s2.length()){
            status = false;
        }else {
            //Convert into character array
 
            char[] s1Array = s1.toLowerCase().toCharArray();
 
            char[] s2Array = s2.toLowerCase().toCharArray();
 
            //Sorting both character array
 
            Arrays.sort(s1Array);
 
            Arrays.sort(s2Array);
 
            //Check if both arrays are equal
 
            status = Arrays.equals(s1Array, s2Array);
        }
        if(status){
            System.out.println(s1+" and "+s2+" are anagrams");
        } else {
            System.out.println(s1+" and "+s2+" are not anagrams");
        }
    }
	
	
}

Method 2 : Without Using Arrays.equals Method

In this approach, we simply traverse any given word and remove each character of first word from the second word. If the words are anagrams string, then the at the end of the loop the second word would be an empty string.

   public static boolean isAnagram(String word, String anagram) {
        boolean isAnagram = false;
        if (word != null && anagram != null && word.length() == anagram.length()) {
            char[] arr = word.toCharArray();
            StringBuilder temp = new StringBuilder(anagram);
            int wordLength = word.length();
            for (char ch : arr) {
                int index = temp.indexOf("" + ch);
                if (index != -1) {
                    temp.deleteCharAt(index);
                }
            }
            isAnagram = temp.toString().isEmpty();
        }
        return isAnagram;
    }

Method 3 : Array Characters Plotting

In this approach, we create an array of size 128 as we use only 128 total character which is used mostly during the program. For each character, we simply increase the count at that character index. The range lies from 0 to 128. Again for the anagram word, we can iterate each character and decrement the count at that character index.

At the end, the array should have all indexes with value as 0 to pass our anagram test.

public static boolean isAnagram(String word, String anagram) {
        boolean isAnagram = true;
        if (word.length() != anagram.length()){
            isAnagram = false;
        }else {
            int[] array = new int[128];
            Arrays.fill(array, 0);
            for (int i = 0; i < word.length(); i++) {
                array[word.charAt(i)] = array[word.charAt(i)] + 1;
            }
            for (int i = 0; i < anagram.length(); i++) {
                array[anagram.charAt(i)] = array[anagram.charAt(i)] - 1;
            }
            for (int i = 0; i < 128; i++) {
                if (array[i] != 0) {
                    isAnagram = false;
                    break;
                }
            }
        }
        return isAnagram;

    }