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Link for the Problem – Java Anagrams – Hacker Rank Solution

Java Anagrams – Hacker Rank 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

• Strings a and b consist 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”.

Java Anagrams – Hacker Rank Solution

mport java.io.*;
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(System.in);
        String a = scan.next();
        String b = scan.next();
        scan.close();
        boolean ret = isAnagram(a, b);
        System.out.println( (ret) ? "Anagrams" : "Not Anagrams" );
    }
}