NumberNames

/**
 *If the numbers 1 to 5 are written out in words: one, two, three, four, five, then there are 3 + 3 + 5 + 4 + 4 = 19 letters used in total.
 *If all the numbers from 1 to 1000 (one thousand) inclusive were written out in words, how many letters would be used?
 *NOTE: Do not count spaces or hyphens. For example, 342 (three hundred and forty-two) contains 23 letters and 115 (one hundred and fifteen) contains 20 letters.
 *The use of "and" when writing out numbers is in compliance with British usage.
 *
 * @author Kushtrim
 * @version 1.01
 */
public class NumberNames

{
    int[] ones =  {0,"one".length(),"two".length(),"three".length(),"four".length(),"five".length(),"six".length(),"seven".length(),"eight".length(),"nine".length()};
    int[] teens = {"ten".length(),"eleven".length(),"twelve".length(),"thirteen".length(),"fourteen".length(),"fifteen".length(),"sixteen".length(),"seventeen".length(),"eighteen".length(),"nineteen".length()};
    int[] tens = {0,0,"twenty".length(),"thirty".length(),"forty".length(),"fifty".length(),"sixty".length(),"seventy".length(),"eighty".length(),"ninety".length()};
    int AND = 3;
    int HUNDRED = 7;

    public static void main(String args[])
    {
        NumberNames llogarit = new NumberNames();

        int shuma=0;

        for (int i = 1 ; i!=1001 ; i++)
        {
            shuma+= llogarit.countLetters(i);
        }
        System.out.println(shuma);

    }
    /**
     * Merr nje numer ( deri ne 1000), dhe kthen numrin e shkronjave qe do duheshin per ta shkruar ( ne gjuhen angleze )
     */
    public int countLetters(int number)
    {
        int n = (number+"").length();
        int letterNumber=0;

        switch(n)
        {
            case 1 :  {
                       letterNumber = processOnes(number);
                       break;
                      }
            case 2 :  {
                        letterNumber = processTens(number);
                        break;
                      }
            case 3 :  {
                        letterNumber = processHundreds(number);
                        break;
                      }
            case 4 :  {
                        letterNumber = "oneThousand".length();
                        break;
                      }
        }
        return letterNumber;
    }
    /**
     * Merr nje numer nje-shifror, dhe kthen numrin e shkronjave qe do duheshin per ta shkruar ( ne gjuhen angleze )
     */
    private int processOnes( int nr)
    {
        int answer = ones[nr];  // numri ne vargun ones, me indeksin nr, paraqet numrin e shkronjave qe do duheshin per ta shkruar atë (nr) ( ne gjuhen angleze )
        return answer;
    }
     /**
     * Merr nje numer prej 10-19 dhe kthen numrin e shkronjave qe do duheshin per ta shkruar ( ne gjuhen angleze )
     */
    private int processTeens( int nr)
    {
        int o = nr%10;           // p.sh.  per 12, llogarit 2
        int answer = teens[o];  // numri ne vargun teens, me indeksin o, paraqet numrin e shkronjave qe do duheshin per ta shkruar  (nr) ( ne gjuhen angleze )
        return answer;
    }
     /**
     * Merr nje numer dy-shifror, dhe kthen numrin e shkronjave qe do duheshin per ta shkruar ( ne gjuhen angleze )
     */
    private int processTens( int nr)
    {
        int t = nr/10;    // p.sh.  per 53, llogarit 5
        int o = nr%10;    // p.sh.  per 53, llogarit 3
        int answer=0;
        if ( t == 1)
        {
            answer = processTeens(nr);  // pra eshte ne intervalin 10-19
        }
        else
        {
            // numri ne vargun tens, me indeksin t, paraqet numrin e shkronjave qe do duheshin per ta shkruar atë (t*10) ( ne gjuhen angleze )
            //p.sh. per 53, tens[5] e kthen gjatesine e "fifty", dhe e therrasim metoden processOnes(3), e cila kthen gjatesine e "three"
            answer = tens[t] + processOnes(o);
        }
        return answer;
    }
     /**
     * Merr nje numer tre-shifror, dhe kthen numrin e shkronjave qe do duheshin per ta shkruar ( ne gjuhen angleze )
     */
    private int processHundreds( int number)
    {
        int h = number/100;    // p.sh.  per 532 kthen 5
        int t = number%100;    // p.sh.  per 532 kthen 32
        int answer=0;
        if (t==0)
        {

            answer = ones[h] + HUNDRED; // per 100,200,...,900  thuhet ONE ( ose TWO,THREE,...,NINE) HUNDRED
        }
        else
        {
            // formati : (ONE,...NINE) HUNDRED AND (ONE, TWO,...,NINETY-NINE)
            answer = ones[h] + HUNDRED + AND + processTens(t);
        }

        return answer;
    }

}