Simple form

Java/Java Anagrams/Solution.java

@@ -29,91 +29,29 @@
 //
 //Not Anagrams
 
-
-
-
-
-import java.io.*;
-import java.util.*;
-
 public class Solution {
-
-	 //My hash map is initialized based on the fact that this is only using English characters as is stated in the problem constraints
 
-	//Overall time complexity: O(n)
-	//Overall space complexity: O(1)
+    static boolean isAnagram(String a, String b){
+        a = a.toUpperCase();
+        b = b.toUpperCase();
+        char[] ar1 = a.toCharArray();
+        java.util.Arrays.sort(ar1);
+        String sorted1 = String.valueOf(ar1); //sort the chars 
 
-	static boolean isAnagram(String a, String b)
-  {
-      boolean anagram = true;
-      Character[] charSet =   {'A','B','C','D','E',
-                            'F','G','H','I','J',
-                            'K','L','M','N','O',
-                            'P','Q','R','S','T',
-                            'U','V','W','X','Y','Z'};
-      HashMap<Character,Integer> ALetterFrequency = new HashMap<Character,Integer>();
-      HashMap<Character,Integer> BLetterFrequency = new HashMap<Character,Integer>();
-      intializeHash(ALetterFrequency, charSet);       //O(n)
-      intializeHash(BLetterFrequency, charSet);       //O(n)
-      a = a.toUpperCase(); //O(n)
-      b = b.toUpperCase(); //O(n)
-
-      //Originally I took a different approach to this problem because I was using substring and it 
-      // runs in O(n) which would make my algorithm run in O(n^2) with constant space, but then I
-      // remembered that charAt() runs in constant time so utilizing that function I was able to
-      // have a time complexity of O(n) and space complexity of O(1)
-
-      for(int i = 0;i <a.length();i++)  //O(n)
-      {
-    	  Character letter = a.charAt(i);   //O(1)
-          Integer frequency = ALetterFrequency.get(letter);
-          ALetterFrequency.put(letter,++frequency);
-      }
-      for(int i = 0;i <b.length();i++)  //O(n)
-      {
-          Character letter = b.charAt(i);   //O(1)
-          Integer frequency = BLetterFrequency.get(letter);
-          BLetterFrequency.put(letter,++frequency);
-      }
-
-
-      //Compare Hash
-      for(Character letter : charSet)
-      {
-          if(!ALetterFrequency.get(letter).equals(BLetterFrequency.get(letter)))
-          {
-              anagram = false;
-          }
-      }
-
-      return anagram;
-  }
-  static void intializeHash(HashMap hash, Character[] set)
-  {
-      for(Character letter : set)
-     {
-          hash.put(letter,0);
-     }
-
-  }
-  static Character[] toCharacterArray(char[] cArray)
-  {
-      Character[] CArray = new Character[cArray.length];
-      for(int i =0; i<cArray.length;i++)
-      {
-          CArray[i] = (Character) cArray[i];
-      }
-      return CArray;
-  }
+        char[] ar2 =  b.toCharArray();
+        java.util.Arrays.sort(ar2);
+        String sorted2 = String.valueOf(ar2);  //sort the char 
 
-	    public static void main(String[] args) {
-
-	        //Scanner scan = new Scanner(System.in);
-	        String a = "anagramm";
-	        String b = "marganaa";
-	        //scan.close();
-	        boolean ret = isAnagram(a, b);
-	        System.out.println( (ret) ? "Anagrams" : "Not Anagrams" );
-	    }
+        return sorted1.equals(sorted2); //Compare the String now 
+    }
+
+    public static void main(String[] args) {
+        //Scanner scan = new Scanner(System.in);
+        String a = "anagramm";
+        String b = "marganaa";
+        //scan.close();
+        boolean ret = isAnagram(a, b);
+        System.out.println( (ret) ? "Anagrams" : "Not Anagrams" );
+    }
 
 }