How to write algorithm for Count Occurrences of Anagrams in java?

Question

Hi I want to write anagram algorithm in java. My requirement is if someone gave a string like this

Input: "aa aa odg dog gdo" its count of anagram count should be 2. Can anyone help me to fix this? I have tried a solution but it is nit working.

public static void main(String[] args) {
        String text = "a c b c run urn urn";
        String word = "urn";
        System.out.print(countAnagrams(text, word));

    }

    static boolean araAnagram(String s1,
                              String s2)
    {
        char[] ch1 = s1.toCharArray();
        char[] ch2 = s2.toCharArray();
        Arrays.sort(ch1);
        Arrays.sort(ch2);
        if (Arrays.equals(ch1, ch2))
            return true;
        else
            return false;
    }

    static int countAnagrams(String text, String word)
    {
        int N = text.length();
        int n = word.length();
        int res = 0;
        for (int i = 0; i <= N - n; i++) {

            String s = text.substring(i, i + n);
            if (araAnagram(word, s))
                res++;
        }
        return res;
    }

This program does not suit with my requirement please help.

Answer 1

Assuming that there is a method to convert a word into an ordered sequence of characters, the number of anagrams may be calculated using Stream API:

static String anagram(String s) {
    char[] arr = s.toCharArray();
    Arrays.sort(arr);
    return new String(arr);
}

static long countAnagrams(String input) {
    if (null == input || input.isEmpty()) {
        return 0;
    }
    return Arrays.stream(input.split("\\s+"))  // Stream of words
            .distinct()  // get unique words and
            // move them into Map<String, List<String>>, where List contains all anagrams contained in the input string
            .collect(Collectors.groupingBy(MyClass::anagram)) 
            .entrySet() // 
            .stream()   // stream of entries 
            .filter(e -> e.getValue().size() > 1) // filter values with at least two anagrams
            .peek(e -> System.out.println(e.getValue())) // debug print of the anagram list
            .count(); // and count them
}

Test:

String[] tests = {
    "aa aa odg dog gdo",
    "cars are very cool so are arcs and my os"
};

Arrays.stream(tests)
      .forEach(s -> System.out.printf("'%s' -> anagram count=%d%n", s, countAnagrams(s)));

Output

[odg, dog, gdo]
'aa aa odg dog gdo' -> anagram count=1
[so, os]
[cars, arcs]
'cars are very cool so are arcs and my os' -> anagram count=2

Answer 2

This can solve.

public static boolean anagrams(String s1, String s2){
    if(!(s1.equalsIgnoreCase(s2))){
        char[] ch1 = s1.toCharArray();
        char[] ch2 = s2.toCharArray();
        Arrays.sort(ch1);
        Arrays.sort(ch2);
        return Arrays.equals(ch1,ch2);
    }
    return false;
}

public static String CountingAnagrams(String str) {
    int res = 0;
    String[] splitStr = str.split("\\s+");
    for(int i = 0 ; i< splitStr.length; i++)
        for (int j = i + 1; j < splitStr.length; j++)
            if (anagrams(splitStr[i], splitStr[j]))
                res++;
    return String.valueOf(res-1);
}