finding if two words are anagrams of each other

Question

I am looking for a method to find if two strings are anagrams of one another.

Ex: string1 - abcde
string2 - abced
Ans = true
Ex: string1 - abcde
string2 - ab         


        

Answer 1

let's take a question: Given two strings s and t, write a function to determine if t is an anagram of s.

For example, s = "anagram", t = "nagaram", return true. s = "rat", t = "car", return false.

Method 1(Using HashMap ):

public class Method1 {

    public static void main(String[] args) {
        String a = "protijayi";
        String b = "jayiproti";
        System.out.println(isAnagram(a, b ));// output => true

    }

    private static boolean isAnagram(String a, String b) {
        Map map = new HashMap<>();
        for( char c : a.toCharArray()) {
            map.put(c,    map.getOrDefault(c, 0 ) + 1 );
        }
        for(char c : b.toCharArray()) {
            int count = map.getOrDefault(c, 0);
            if(count  == 0 ) {return false ; }
            else {map.put(c, count - 1 ) ; }
        }

        return true;
    }

}

Method 2 :

public class Method2 {
public static void main(String[] args) {
    String a = "protijayi";
    String b = "jayiproti";


    System.out.println(isAnagram(a, b));// output=> true
}

private static boolean isAnagram(String a, String b) {


    int[] alphabet = new int[26];
    for(int i = 0 ; i < a.length() ;i++) {
         alphabet[a.charAt(i) - 'a']++ ;
    }
    for (int i = 0; i < b.length(); i++) {
         alphabet[b.charAt(i) - 'a']-- ;
    }

    for(  int w :  alphabet ) {
         if(w != 0 ) {return false;}
    }
    return true;

}
}

Method 3 :

public class Method3 {
public static void main(String[] args) {
    String a = "protijayi";
    String b = "jayiproti";


    System.out.println(isAnagram(a, b ));// output => true
}

private static boolean isAnagram(String a, String b) {
    char[] ca = a.toCharArray() ;
    char[] cb = b.toCharArray();
    Arrays.sort(   ca     );

    Arrays.sort(   cb        );
    return Arrays.equals(ca , cb );
}
}

Method 4 :

public class AnagramsOrNot {
    public static void main(String[] args) {
        String a = "Protijayi";
        String b = "jayiProti";
        isAnagram(a, b);
    }

    private static void isAnagram(String a, String b) {
        Map map = new LinkedHashMap<>();

        a.codePoints().forEach(code -> map.put(code, map.getOrDefault(code, 0) + 1));
        System.out.println(map);
        b.codePoints().forEach(code -> map.put(code, map.getOrDefault(code, 0) - 1));
        System.out.println(map);
        if (map.values().contains(0)) {
            System.out.println("Anagrams");
        } else {
            System.out.println("Not Anagrams");
        }
    }
}

In Python:

def areAnagram(a, b):
    if len(a) != len(b): return False
    count1 = [0] * 256
    count2 = [0] * 256
    for i in a:count1[ord(i)] += 1
    for i in b:count2[ord(i)] += 1

    for i in range(256):
        if(count1[i] != count2[i]):return False    

    return True


str1 = "Giniiii"
str2 = "Protijayi"
print(areAnagram(str1, str2))

Let's take another famous Interview Question: Group the Anagrams from a given String:

public class GroupAnagrams {
    public static void main(String[] args) {
        String a = "Gini Gina Protijayi iGin aGin jayiProti Soudipta";
        Map> map = Arrays.stream(a.split(" ")).collect(Collectors.groupingBy(GroupAnagrams::sortedString));
        System.out.println("MAP => " + map);
        map.forEach((k,v) -> System.out.println(k +" and the anagrams are =>" + v ));
        /*
         Look at the Map output:
        MAP => {Giin=[Gini, iGin], Paiijorty=[Protijayi, jayiProti], Sadioptu=[Soudipta], Gain=[Gina, aGin]}
        As we can see, there are multiple Lists. Hence, we have to use a flatMap(List::stream)
        Now, Look at the output:
        Paiijorty and the anagrams are =>[Protijayi, jayiProti]

        Now, look at this output:
        Sadioptu and the anagrams are =>[Soudipta]
        List contains only word. No anagrams.
        That means we have to work with map.values(). List contains all the anagrams.


        */
        String stringFromMapHavingListofLists = map.values().stream().flatMap(List::stream).collect(Collectors.joining(" "));
        System.out.println(stringFromMapHavingListofLists);
    }

    public static String sortedString(String a) {
        String sortedString = a.chars().sorted()
                .collect(StringBuilder::new, StringBuilder::appendCodePoint, StringBuilder::append).toString();

        return sortedString;

    }

    /*
     * The output : Gini iGin Protijayi jayiProti Soudipta Gina aGin
     * All the anagrams are side by side.
     */
}