Letters Frequency Counter

[alfonz jan frithz]

I have several solution for this case actually. I just want to ask if this code "sexy" enough.
If you have suggestion, i really appreciate it...

import java.util.Scanner;

public class LettersCounter {
  public static void main(String[]args){
    Scanner sc = new Scanner(System.in);
    String s = "";
    int counter = 0;

    System.out.print("Please insert some text: ");
    String src = sc.next().trim();
    char [] c = src.toCharArray();
    
    for(int i=0;i<c.length;i++) for(int j=i+1;j<c.length;j++) if(c[i] == c[j]) c[j] = '-';
    
    for(int i=0;i<c.length;i++){
      if(c[i]!='-'){
        counter++;
        s+=String.valueOf(c[i]);
      }
    }

    int f[] = new int[counter];

    for(int i=0;i<s.length();i++){
      for(int j=0;j<c.length;j++){
        System.out.println(c[j]+" "+s.charAt(i));
        if(src.charAt(j)==s.charAt(i)) f[i]++;
      }
    }

    for(int i=0;i<s.length();i++) System.out.println(s.charAt(i)+" "+f[i]);
    
  }
}

Thanks...

[Miklos Szeles]

Hi Alfonz,

I don't think this algorithm is too "sexy". It's very inefficient. You use a for loop in a for loop which runs through the text again and again and you do other unnecesseary operations.
You can do the counting part with one for loop and a Map. You run through the text with the loop and for every character you have 2 options:
1, Character not in the map, you put it into the map with a value of 1.
2, The Map contains the chraracter: You increment the stored value by one.
From the map you know the:
-number of different characters
-which are these characters
-how many of them you have
So you have all the necesseary information.
I hope it helps.

[alfonz jan frithz]

thanks for the solution... its cool man...