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How NetBeans helped in 100th day of school celebration ?

Posted by arungupta on February 3, 2008 at 5:13 PM PST

My friend's daughter got a family assignment from school to commemorate 100th
day in school. If each letter in the English alphabet is assigned a weight
corresponding to it's position (e.g. "a" is 1, "b" is 2, ... "z" is 26) then she
was supposed to collect 10 words whose letters sum total to 100.

They spent last
1.5 days and found only 3 words and were discussing the problem at the lunch
table earlier today. Here is a small piece of code that I wrote to help them



import java.util.Arrays;

import java.util.StringTokenizer;

public class Main {

    private static final char[] WEIGHT = {

        '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'



    * @param args the command line arguments


    public static void main(String[] args) throws Exception {

        if (args.length == 0) {

System.out.println("At least one argument required");



        BufferedReader in = new
BufferedReader(new FileReader(args[0]));

        String string = in.readLine();

        while (string != null) {

StringTokenizer tokenizer = new StringTokenizer(string);

            while (tokenizer.hasMoreTokens())

String word = tokenizer.nextToken();

if (wordStrength(word) == 100)

System.out.println("100 letters old: " + word);


            string =



    static int wordStrength(String word) {

        int strength = 0;

        for (int i=0; i<word.length(); i++) {

charStrength = Arrays.binarySearch(WEIGHT, word.charAt(i));

            if (charStrength
>= 0)

strength += charStrength+1;






This application was easily created using
NetBeans 6
. And then I copy/pasted articles from my favorite news sites in a
file, passed it as command-line argument by right-clicking the project,
selecting "Properties" and specifying the file name in "Arguments".

Hit F6, and voila got more than 10 words :)



Related Topics >>


You could also take the entire aspell dictionary and search that.

Also I think you could eliminate the binary search,by casting a char to an int.
int base = (int)'a' - 1; System.out.println('a' - base); System.out.println('b' - base); System.out.println('c' - base); ...output... 1 2 3

Sure it's horribly wrong for internationalization but you are searching only English worlds. I would suspect the casting and subtraction would be faster then the binary search. It's a very cool little problem.

What does this have to do with NetBeans?

Quick code completion and suggestions can be found in any other IDE of matching strength (which would be at least Eclipse, IDEA and JDeveloper).

You can write this code any IDE or event notepad/vi but I used NetBeans. Quick code completion and suggestions reduced the time to author this code.