Java Anagrams HackerRank 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 .
The second line contains a string denoting .
Constraints
- Strings and consist of English alphabetic characters.
- The comparison should NOT be case sensitive.
Output Format
Print "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) {
// Complete the function
int arr[]=new int[123];
boolean isAnagram=true;
for(char c:a.toCharArray())
{ int index=(int)c;
if(65<=index && index<=90)
index+=32;
arr[index]++;
}
for(char c:b.toCharArray()) // O(n) Time Complexity
{ int index=(int)c;
if(65<=index && index<=90)
index+=32;
arr[index]--;
}
for(int i=0;i<123;i++)
{ if(arr[i]!=0)
isAnagram=false;
}
return isAnagram;
}
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" );
}
}
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in Java.
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