Java Anagrams | HackerRank Solution

Hello coders, today we are going to solve 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 CAT, ACT, tac, TCA, aTC, 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

Returns

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.

Constraints

• 1 <= length(a), length(b) <= 50
• Strings a and b consists of English alphabetic characters.
• The comparison should NOT be case sensitive.

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 – Java Anagrams

import java.util.Scanner;

public class Solution {

    static boolean isAnagram(String a, String b) {
        // Complete the function
        String s1 = a;
        String s2 = b;
        s1=s1.toLowerCase();
        s2=s2.toLowerCase();
        
        if(s1.length()==s2.length())

        {
            int[] arr = new int[256];
            int[] brr = new int[256];
            for (int i = 0; i < s1.length(); i++) {
                arr[(int) s1.charAt(i)] += 1;
                brr[(int) s2.charAt(i)] += 1;
            }
            for (int i = 0; i < 256; i++) {
                if (arr[i] != brr[i])
                    return false;

            }
            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" );
    }
}

Disclaimer: The above Problem ( Java Anagrams ) is generated by Hacker Rank but the Solution is Provided by CodingBroz. This tutorial is only for Educational and Learning Purpose.