// Efficient : Time Complexity : O(n)
 
import java.util.*;
import java.io.*;
  
class GFG { 
    
    static final int CHAR=256;
        
    static boolean areAnagram(String s1, String s2) 
    { 
       if (s1.length() != s2.length()) 
            return false; 
  
        int[] count=new int[CHAR];
        for(int i=0;i<s1.length();i++){
            count[s1.charAt(i)]++;
            count[s2.charAt(i)]--;
        }
    
        for(int i=0;i<CHAR;i++){
            if(count[i]!=0)return false;
        }
        return true;
    }
  
    public static void main(String args[]) 
    { 
        String str1 = "abaac"; 
        String str2 = "aacba";  
        if (areAnagram(str1, str2)) 
            System.out.println("The two strings are" + " anagram of each other"); 
        else
            System.out.println("The two strings are not" + " anagram of each other"); 
    } 
} 
 
 

// Naive : Time Complexity : Θ(nlogn)
 
    static boolean areAnagram(String s1, String s2) 
    { 
       
        if (s1.length() != s2.length()) 
            return false; 
  
       char a1[]=s1.toCharArray();
        Arrays.sort(a1);
        s1=new String(a1);
        char a2[]=s2.toCharArray();
        Arrays.sort(a2);
        s2=new String(a2);
        
        return s1.equals(s2);
    }