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

Code:

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" );

}

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

Java Anagrams – Hacker Rank Solution

Problem :

Java Anagrams – Hacker Rank Solution

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