Help with a Programming assignment

[Deron Brown]

public class Assignment9
{
	public static void main ( String [] args )
	{
		String str = Input.getString("Please type in a word");
		System.out.println("The word is " + str);

		int [] lettercount = new int [26];	
		
		for ( int k = 0; k < str.length(); k++ )
		{
			if (Character.isLetter(str.charAt(k)))
			{ 
				char [] chars = new char [str.charAt(k)];
				lettercount[chars[k] - 'a']++;			
			}

		}
		
		System.out.println( "A = " + lettercount[0] );
		System.out.println( "B = " + lettercount[1] );
		System.out.println( "C = " + lettercount[2] );
		System.out.println( "D = " + lettercount[3] );
		System.out.println( "E = " + lettercount[4] );
		System.out.println( "F = " + lettercount[5] );
		System.out.println( "G = " + lettercount[6] );
		System.out.println( "H = " + lettercount[7] );
		System.out.println( "I = " + lettercount[8] );
		System.out.println( "J = " + lettercount[9] );
		System.out.println( "K = " + lettercount[10] );
		System.out.println( "L = " + lettercount[11] );
		System.out.println( "M = " + lettercount[12] );
		System.out.println( "N = " + lettercount[13] );
		System.out.println( "O = " + lettercount[14] );
		System.out.println( "P = " + lettercount[15] );
		System.out.println( "Q = " + lettercount[16] );
		System.out.println( "R = " + lettercount[17] );
		System.out.println( "S = " + lettercount[18] );
		System.out.println( "T = " + lettercount[19] );
		System.out.println( "U = " + lettercount[20] );
		System.out.println( "V = " + lettercount[21] );
		System.out.println( "W = " + lettercount[22] );
		System.out.println( "X = " + lettercount[23] );
		System.out.println( "Y = " + lettercount[24] );
		System.out.println( "Z = " + lettercount[25] );
		
	}
}

Just another update same result just added in the other letters

[Rico Felix]

I'm glad that you are at least thinking about different solutions...

Now the question is... what is the meaning of this -> char [] chars = new char [str.charAt(k)];

Explain what you're doing in that line of code

[Deron Brown]

Rico Felix wrote:I'm glad that you are at least thinking about different solutions...

Now the question is... what is the meaning of this -> char [] chars = new char [str.charAt(k)];

Explain what you're doing in that line of code

Honestly I have no idea I think I was trying to store the characters from charAt into an array and use that

[Rico Felix]

What that line does is create a new array of type char with the size of the integer representation of the character returned from the method invocation... doesn't do much good for what you are trying to achieve


Look at this section of code carefully:

for ( int k = 0; k < str.length(); k++ )
{
    if (Character.isLetter(str.charAt(k)))
    { 
        // void	          
    }

}

If we carefully study what it does then we can probably fill in the void within the if statement
- We are looping through the length of the string... that mean if we have the string "Java" we are looping 4 times since it is made up of four characters
- For each iteration we are checking to see if the character in the string at index k is a letter... if its a letter we want to know what letter it is so we can increment the slot in the array that's keeping our running count of occurrences

We have the following array to keep our running count

int[] arr = new int[26];

This was set up because we know there are 26 letters in the alphabet so incrementing index 0 will say that we found an 'A' or 'a'... incrementing index 1 will say that we found a 'B' or 'b' etc...
In order to know which index in the array to increment within the if statement we came up with an algorithm that say since we are in the if statement its a letter so get the letter, transform it to lowercase and subtract 'a' from it to get the array subscript and increment

This is what you need to do in words... now transform it to working Java code

[Deron Brown]

Rico Felix wrote:What that line does is create a new array of type char with the size of the integer representation of the character returned from the method invocation... doesn't do much good for what you are trying to achieve


Look at this section of code carefully:

for ( int k = 0; k < str.length(); k++ )
{
    if (Character.isLetter(str.charAt(k)))
    { 
        // void	          
    }

}

If we carefully study what it does then we can probably fill in the void within the if statement
- We are looping through the length of the string... that mean if we have the string "Java" we are looping 4 times since it is made up of four characters
- For each iteration we are checking to see if the character in the string at index k is a letter... if its a letter we want to know what letter it is so we can increment the slot in the array that's keeping our running count of occurrences

We have the following array to keep our running count

int[] arr = new int[26];

This was set up because we know there are 26 letters in the alphabet so incrementing index 0 will say that we found an 'A' or 'a'... incrementing index 1 will say that we found a 'B' or 'b' etc...
In order to know which index in the array to increment within the if statement we came up with an algorithm that say since we are in the if statement its a letter so get the letter, transform it to lowercase and subtract 'a' from it to get the array subscript and increment

This is what you need to do in words... now transform it to working Java code

Jeez after two days of hard work I've got it working thanks to everyone who responded and tried to help me solve my problem! Special thanks to Rico Felix for being there and helping me through it one step at a time I've got the working code

public class Assignment9
{
	public static void main ( String [] args )
	{
		String str = Input.getString("Please type in a word");
		System.out.println("The word is " + str);

		int [] lettercount = new int [26];	
		
		for ( int k = 0; k < str.length(); k++ )
		{
			if (Character.isLetter(str.charAt(k)))
			{ 
				str.toLowerCase(); 
				lettercount[str.charAt(k) - 'a']++;			
			}

		}
		
		System.out.println( "A = " + lettercount[0] );
		System.out.println( "B = " + lettercount[1] );
		System.out.println( "C = " + lettercount[2] );
		System.out.println( "D = " + lettercount[3] );
		System.out.println( "E = " + lettercount[4] );
		System.out.println( "F = " + lettercount[5] );
		System.out.println( "G = " + lettercount[6] );
		System.out.println( "H = " + lettercount[7] );
		System.out.println( "I = " + lettercount[8] );
		System.out.println( "J = " + lettercount[9] );
		System.out.println( "K = " + lettercount[10] );
		System.out.println( "L = " + lettercount[11] );
		System.out.println( "M = " + lettercount[12] );
		System.out.println( "N = " + lettercount[13] );
		System.out.println( "O = " + lettercount[14] );
		System.out.println( "P = " + lettercount[15] );
		System.out.println( "Q = " + lettercount[16] );
		System.out.println( "R = " + lettercount[17] );
		System.out.println( "S = " + lettercount[18] );
		System.out.println( "T = " + lettercount[19] );
		System.out.println( "U = " + lettercount[20] );
		System.out.println( "V = " + lettercount[21] );
		System.out.println( "W = " + lettercount[22] );
		System.out.println( "X = " + lettercount[23] );
		System.out.println( "Y = " + lettercount[24] );
		System.out.println( "Z = " + lettercount[25] );
		
	}
}

[Rico Felix]

Although your code works you aren't finished yet... simply because the method you chose to print the information is not good practice... it should be done using a loop, that's why they were created to avoid repetitive typing and duplication

[Rico Felix]

Since you mentioned its two days you were working on this I think you won't want any more to do with this nuisance...

Here is a gift of the my version:

/**
 * This utility component counts letter occurrences in a string
 */
public class LetterOccurrenceCount {

	/**
	 * Print letter occurrences in a string
	 * @param string String to be proccessed
	 */
	public static void printLetterOccurrences(String string) {

		int[] information = countLetterOccurrences(string);

		/* +++++++++++++++++++++++++++++++++
		 * Values to start printing process
		 * ================================*/
		char initialCapitalLetter = 'A';
		char initialCommonLetter = 'a';

		/* ++++++++++++++++++++++++++++++++++++++++++
		 * Loop through the information and print it
		 * =========================================*/
		for (int i = 0; i < information.length; i++) {
			System.out.println(
				((char)initialCapitalLetter++)
				+ ""
				+ ((char)initialCommonLetter++)
				+ " = "
				+ information[i]);
		}

	}

	/**
	 * Method to use as a test driver for this
	 * component
	 */
	public static void test() {

		/* ++++++++++++++++++++
		 * Get input from user
		 * ===================*/
		String textInput = System
							.console()
							.readLine("%s", "Enter some text: ");

		printLetterOccurrences(textInput);

	}

	/* +++++++++++++++++++++++++++++++++++++++
	 * Method for counting letter occurrences
	 * ======================================*/
	private static int[] countLetterOccurrences(String text) {

		int[] letterRegistry = new int[26]; // to store letter occurrences
		int index = 0;

		/* ++++++++++++++++++++++++++++++++++++++++
		 * Loop until all characters are processed
		 * in the input string
		 * =======================================*/
		while(index < text.length()) {

			/* +++++++++++++++++++++++++++++++++++++++
			 * Check if current character is a letter
			 * If yes add its occurrence to the correct
			 * position in the registry
			 * ======================================*/
			if (Character.isLetter(text.charAt(index))) {

				/* ++++++++++++++++++++++++++++++++
				 * Increment correct registry slot
				 * according to letter found
				 * ===============================*/
				letterRegistry[Character.toLowerCase(
								text.charAt(index)) - 'a']++;
			}

			index++;
		}

		return (letterRegistry);
	}

	/* +++++++++++++++++++++++++++++++++++++++++++
	 * Stop clients from creating objects of this
	 * component
	 * ==========================================*/
	private LetterOccurrenceCount() {

	}

}

public class Test {
	public static void main(String[] args) {
		if (args.length == 0) {
			System.out
				.println("Usage: java Test \"argument\"");
			System.exit(0x0);
		}
		LetterOccurrenceCount
			.printLetterOccurrences(args[0]);
	}
}

[Campbell Ritchie]

Deron Brown wrote: . . .

public class Assignment9
{
//...
		System.out.println( "A = " + lettercount[0] );
		System.out.println( "B = " + lettercount[1] );
		System.out.println( "C = " + lettercount[2] );
		System.out.println( "D = " + lettercount[3] );
		System.out.println( "E = " + lettercount[4] );
		System.out.println( "F = " + lettercount[5] );
		System.out.println( "G = " + lettercount[6] );
		System.out.println( "H = " + lettercount[7] );
		System.out.println( "I = " + lettercount[8] );
		System.out.println( "J = " + lettercount[9] );
		System.out.println( "K = " + lettercount[10] );
		System.out.println( "L = " + lettercount[11] );
		System.out.println( "M = " + lettercount[12] );
		System.out.println( "N = " + lettercount[13] );
		System.out.println( "O = " + lettercount[14] );
		System.out.println( "P = " + lettercount[15] );
		System.out.println( "Q = " + lettercount[16] );
		System.out.println( "R = " + lettercount[17] );
		System.out.println( "S = " + lettercount[18] );
		System.out.println( "T = " + lettercount[19] );
		System.out.println( "U = " + lettercount[20] );
		System.out.println( "V = " + lettercount[21] );
		System.out.println( "W = " + lettercount[22] );
		System.out.println( "X = " + lettercount[23] );
		System.out.println( "Y = " + lettercount[24] );
		System.out.println( "Z = " + lettercount[25] );
		
	}
}

No. That is dreadful. You have written the same code twenty‑six times. That needs to go in a loop.