Hackerrank Java Anagrams Solution

Hackerrank Java Anagrams Solution

import java.util.Scanner;
public class Solution {
    static boolean isAnagram(String a, String b) {
       
//      // once you declare a.toUppercase you should assign it to a. you cannot define it as just a.toUppercase... 
//      //I solved it with the long way however I could put a and b in a character array and then use  Arrays.sort(arrayname). after this steps convert them to string and check if they are equel.
        a=a.toUpperCase();
        b=b.toUpperCase();
        boolean ret = false;
        StringBuilder c= new StringBuilder(b);

        if(a.length()==b.length()){
            for(int i=0; i<a.length();i++){
                for(int j=0; j<c.length();j++){
                    if(a.charAt(i)==c.charAt(j)){
                            c.deleteCharAt(j);
                        if(i==a.length()-1 && c.length()==0){
                            ret=true;
                            break;
                        }
                        break;

                    }

                }
            }
        }return ret;
   
   }
        
         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" );
    }
}

Nice one, thanks a lot Abdullah..................

thx

That's slow algorithm

import java.util.Scanner;

public class Solution {

static boolean isAnagram(String c, String d) {
  
  String a=c.toLowerCase();
  String b=d.toLowerCase();
  int count=0;
  int count2=0;

  if(a.length()!=b.length())
  {
      return false;
  }

  for(int i=0; i<a.length(); i++)
  {
      char ch=a.charAt(i);
      for(int j=0; j<a.length(); j++)
      {
          if(a.charAt(j)==ch)
          count++;
          if(b.charAt(j)==ch)
          count2++;
      }
      if(count!=count2)
      {
      return false; 
      }
  }
  return true;
}

public static void main(String[] args) {

what's the use of string builder? can anyone please explain it to me?

StringBuilder class can be used when you want to modify a string without creating a new object. For example, using the StringBuilder class can boost performance when concatenating many strings together in a loop
StringBuilder objects are like String objects, except that they can be modified. Internally, these objects are treated like variable-length arrays that contain a sequence of characters. At any point, the length and content of the sequence can be changed through method invocations.

Strings should always be used unless string builders offer an advantage in terms of simpler code (see the sample program at the end of this section) or better performance. For example, if you need to concatenate a large number of strings, appending to a StringBuilder object is more efficient.

Length and Capacity
The StringBuilder class, like the String class, has a length() method that returns the length of the character sequence in the builder.

Unlike strings, every string builder also has a capacity, the number of character spaces that have been allocated. The capacity, which is returned by the capacity() method, is always greater than or equal to the length (usually greater than) and will automatically expand as necessary to accommodate additions to the string builder.

@anushree71199

import java.io.;
import java.util.;

public class Solution {
public static String lvalue(String s){
char str[]=new char[s.length()];
str=s.toCharArray();
char temp;
for(int i=0;i<s.length();i++){
for(int j=0;j<s.length()-1;j++){
temp=str[j+1];
if(str[j]>str[j+1]){
str[j+1]=str[j];
str[j]=temp;
}
}
}
String ss = new String(str);
return ss;
}
public static void main(String[] args) {
String str1,str2;
Scanner in = new Scanner(System.in);
str1=in.next();
str2=in.next();
str1=str1.toLowerCase();
str2=str2.toLowerCase();
if(str1.length()==str2.length()){
str1=lvalue(str1);
str2=lvalue(str2);
if(str1.equals(str2))
System.out.println("Anagrams");
else
System.out.println("Not Anagrams");
}
else
System.out.println("Not Anagrams");
}
}

A much more simple solution:

private static boolean isAnagram(String a, String b) {
    if (a.length() != b.length()) {
        return false;
    }
    a = a.toLowerCase();
    b = b.toLowerCase();
    
    for (int i = 0; i < b.length(); i++) {
        a = a.replaceFirst(String.valueOf(b.charAt(i)), "");
    }

    return a.isEmpty() ? true : false;
}

What about this solution guys?

package chapter01.exercise_18;

public class CheckTwoStringsAnagram {

    public static void main(String[] args) {
        var one = "geeksforgeeks";
        var two = "forgeeksgeeks";
        var areAnagrams = checkIfTwoStringsAreAnagrams(one, two);
        System.out.format("The strings \"%s\" and \"%s\" are anagrams? %b", one, two, areAnagrams);
    }

    private static boolean checkIfTwoStringsAreAnagrams(final String one, final String two) {
        var oneWithoutSpacesAndLowerCase = one.replaceAll("\\s", "").toLowerCase();
        var twoWithoutSpacesAndLowerCase = two.replaceAll("\\s", "").toLowerCase();
        var countOneCharacters = oneWithoutSpacesAndLowerCase.chars().sum();
        var countTwoCharacters = twoWithoutSpacesAndLowerCase.chars().sum();
        return !oneWithoutSpacesAndLowerCase.equals(twoWithoutSpacesAndLowerCase)
                && countOneCharacters == countTwoCharacters;
    }

}

A much more simple solution:

private static boolean isAnagram(String a, String b) {
    if (a.length() != b.length()) {
        return false;
    }
    a = a.toLowerCase();
    b = b.toLowerCase();
    
    for (int i = 0; i < b.length(); i++) {
        a = a.replaceFirst(String.valueOf(b.charAt(i)), "");
    }

    return a.isEmpty() ? true : false;
}

Is more simple than everything 👯

static boolean isAnagram(String a, String b) {
        // Complete the function
        a = a.toLowerCase();
        b = b.toLowerCase();
        int aVal = 0, bVal = 0;
        
        
        if ( a.length() != b.length() )
            return false;
        for ( int i = 0; i < a.length(); i++ ) {
            aVal += (int)a.charAt(i)*(int)a.charAt(i);
            bVal += (int)b.charAt(i)*(int)b.charAt(i);
        }
        
        return aVal == bVal ? true : false;
    }

seems this is more simpler...

import java.util.Scanner;

public class Solution {

static boolean isAnagram(String a, String b) {
    // Complete the function
    if(a.length() == b.length()){
        
    
        a=a.toUpperCase();
        b=b.toUpperCase();
        char[] ch1 = a.toCharArray();
        char[] ch2 = b.toCharArray();
        char temp;
        for(int i=0;i<ch1.length;i++){
            for(int j=i+1;j<ch1.length;j++){
                if(ch1[i]>ch1[j]){
                    temp =  ch1[i];
                    ch1[i] =ch1[j];
                    ch1[j] = temp; 
                }
            }
        }
        for(int i=0;i<ch2.length;i++){
            for(int j=i+1;j<ch2.length;j++){
                if(ch2[i]>ch2[j]){
                    temp =  ch2[i];
                    ch2[i] =ch2[j];
                    ch2[j] = temp; 
                }
            }
        }
        //System.out.println(ch1);
        //System.out.println(ch2);
        
        //int[] charcounta = new int[ch1.length];
        for (int i = ch1.length - 1; i >= 0; i--)   
        {  
            if(ch1[i]!=ch2[i]){
                return false;
            }else{
                continue;
            }
            
        }  
        return true;
    }
    return false;
}

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" );
}

}