Java Anagrams Hacker Rank Solution

 Problem:-

Two strings,  and , are called anagrams if they contain all the same characters in the same frequencies. For example, the anagrams of CAT are CAT, ACT, TAC, TCA, ATC, and CTA.Complete the function in the editor. If  and  are case-insensitive anagrams, print “Anagrams”; otherwise, print “Not Anagrams” instead.

Input Format

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

Constraints

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

Output FormatPrint “Anagrams” if  and  are case-insensitive anagrams of each other; otherwise, print “Not Anagrams” instead.

Sample Input 0

anagram
margana

Sample Output 0

Anagrams

Explanation 0

Character	Frequency: anagram	Frequency: margana
A or a	3	3
G or g	1	1
N or n	1	1
M or m	1	1
R or r	1	1

The two strings contain all the same letters in the same frequencies, so we print “Anagrams”.Sample Input 1

anagramm
marganaa

Sample Output 1

Not Anagrams

Explanation 1

Character	Frequency: anagramm	Frequency: marganaa
A or a	3	4
G or g	1	1
N or n	1	1
M or m	2	1
R or r	1	1

The two strings don’t contain the same number of a‘s and m‘s, so we print “Not Anagrams”.Sample Input 2

Hello
hello

Sample Output 2

Anagrams

Explanation 2

Character	Frequency: Hello	Frequency: hello
E or e	1	1
H or h	1	1
L or l	2	2
O or o	1	1

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

Solution:-

import java.util.Scanner;



public class Solution {



  static boolean isAnagram(String a, String b) {



        if (a.length() != b.length()) return false;


        


        a = a.toLowerCase();


        b = b.toLowerCase();


        


        int [] temp = new int [Character.MAX_VALUE]; 


    //sry, too lazy to calculate ASCII idx of a-z rel. i :)


        


        int summ = 0;



        for (int i = 0; i < a.length(); i++){


            summ += ++temp [a.charAt(i)] <= 0 ? -1 : 1;


            summ += --temp [b.charAt(i)] >= 0 ? -1 : 1;


        }



        return summ == 0;


        


}


    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" );


    }


}