HackerRank Anagram program in JAVA is running fine on my Eclipse but not on HackerRank platform

I am trying out a HackerRank problem in JAVA. It is running fine on my Eclipse but it is not giving the expected output on HackerRank platform. The problem is to check whether two strings are anagrams of each other or not (ignoring their case). Link to the problem: https://www.hackerrank.com/challenges/java-anagrams/problem?isFullScreen=true

Below is the code:

import java.util.Scanner;

public class Solution {

    static boolean isAnagram(String a, String b) {
        if(a.length()!=b.length())
        return false;
        char c1[]=a.toCharArray();
        char c2[]=b.toCharArray();
        java.util.Arrays.sort(c1);
        java.util.Arrays.sort(c2);
        a=String.valueOf(c1);
        b=String.valueOf(c2);

        if(a.equalsIgnoreCase(b))
        return true;
        else
        return false;
    }

  public static void main(String[] args) {
    
        Scanner scan = new Scanner(System.in);
        String a = scan.next();
        String b = scan.next();
        scan.close();
        boolean ret = isAnagram(a, b);
        System.out.println( (ret) ? "Anagrams" : "Not Anagrams" );
    }
}

It is passing two test cases but not the third one in which String a="Hello" and String b="hello".
But it is passing this test case on eclipse.

Please advise.

At the end of the sorting your arrays contains,

    char c1[]=a.toCharArray();
    char c2[]=b.toCharArray();
    java.util.Arrays.sort(c1); // c1 = ['H','e','l','l','o']
    java.util.Arrays.sort(c2); // c2 = ['e','h','l','l','o']

After this step when you convert this back to String,

    a=String.valueOf(c1); //a = "Hello"
    b=String.valueOf(c2); //b = "ehllo"

Hence it return false when you compare the a and b.

Solution: you might want to convert the strings to lowercase/uppercase before you sort them by using (toLowerCase() or toUpperCase()).

To solve this problem we have to create an int array for character frequency of size 26 ( 26 size is for every alphabet) which is initialized to zero so that we have 26 zeros in the array.

int[] frequency = new int[26];

then we will loop through first string creating an int index which will count the frequency of each alphabet in the string and increment the index of the frequency array representing that particular alphabet for example if the string is CAT the frequency[0], frequency[2]and frequency[19] will be incremented as 'C', 'A', and 'T' are represented by 2,0 and 19 indexes respectively.

 for(int i=0;i<a.length();i++){
        int index = a.charAt(i)-'a';
        frequency[index]++;
    }

now we will loop through second string the same way we did with first-string but in this case, rather then incrementing the frequency array we will decrement the frequency.

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

finally, we will check the frequency array if after performing the above operation all the indexes of frequency array come to zero it means that strings are anagram as the alphabets found in the first string which increment the frequency when found in the second string will decrease the frequency.

 for(int i=0;i<26;i++){
        if(frequency[i] != 0)
          return false;
    }

don't forget to convert string to lowercase before starting this otherwise, it shows runtime error.

     a = a.toLowerCase();
     b = b.toLowerCase();

for complete solution visit https://github.com/Gursimir/Java-Practice/blob/master/anagram.java