Problem:-
Solution:-
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) {
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" );
}
}
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