Java Anagrams HackerRank Solution

Hello Programmers, In this post, you will know how to solve the Java Anagrams HackerRank Solution. This problem is a part of the HackerRank Java Programming Series.

Java Anagrams HackerRank Solution
Java Anagrams HackerRank Solutions

One more thing to add, don’t directly look for the solutions, first try to solve the problems of Hackerrank by yourself. If you find any difficulty after trying several times, then you can look for solutions.

Java Anagrams HackerRank Solution

Problem :

Two strings, a and b, are called anagrams if they contain all the same characters in the same frequencies. For this challenge, the test is not case-sensitive. For example, the anagrams of CAT are CATACTtacTCAaTC, and CtA.

Function Description

Complete the isAnagram function in the editor.

isAnagram has the following parameters:

  • string a: the first string
  • string b: the second string


  • boolean: If a and b are case-insensitive anagrams, return true. Otherwise, return false.

Input Format

The first line contains a string a.
The second line contains a string b.


  • Strings a and b consist of English alphabetic characters.
  • The comparison should NOT be case sensitive.

Sample Input 0


Sample Output 0


Explanation 0

CharacterFrequency: anagramFrequency: margana
A or a33
G or g11
N or n11
M or m11
R or r11

The two strings contain all the same letters in the same frequencies, so we print “Anagrams”.

Sample Input 1


Sample Output 1

Not Anagrams

Explanation 1

CharacterFrequency: anagrammFrequency: marganaa
A or a34
G or g11
N or n11
M or m21
R or r11

The two strings don’t contain the same number of a‘s and m‘s, so we print “Not Anagrams”.

Sample Input 2


Sample Output 2


Explanation 2

CharacterFrequency: HelloFrequency: hello
E or e11
H or h11
L or l22
O or o11

The two strings contain all the same letters in the same frequencies, so we print “Anagrams”.

Java Anagrams HackerRank Solution

import java.util.*;
public class Solution {
    static boolean isAnagram(String a, String b) {
        if ((a == null || b == null) || (a.length() != b.length())) {
            return false;
        final char[] ARRAY_A = a.toUpperCase().toCharArray();
        final Map map = new HashMap<>();
        for (int i = 0; i < ARRAY_A.length; i ++) {
            if (map.containsKey(ARRAY_A[i])) {
                map.put(ARRAY_A[i], (map.get(ARRAY_A[i]) + 1));
            } else {
                map.put(ARRAY_A[i], 1);
        final char[] ARRAY_B = b.toUpperCase().toCharArray();
        for (int i = 0; i < ARRAY_A.length; i ++) {
            if (map.containsKey(ARRAY_B[i])) {
                Integer count = map.get(ARRAY_B[i]);
                if (count == 0) {
                    return false;
                } else {
                    count --;
                    map.put(ARRAY_B[i], count);
            } else {
                return false;
        return true;
    public static void main(String[] args) {
        Scanner scan = new Scanner(;
        String a =;
        String b =;
        boolean ret = isAnagram(a, b);
        System.out.println( (ret) ? "Anagrams" : "Not Anagrams" );

Disclaimer: The above Problem (Java Anagrams) is generated by Hackerrank but the Solution is Provided by BrokenProgrammers. This tutorial is only for Educational and Learning purposes.

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