Thread: Pangram Question (All the letters of the alphabet in a sentence)

I've searched through the forum to see if this question has been posted previously but I can't find it so apologies in advance if I missed it...

What I am trying to do is to run the program to check through each letter of the sentence and as it does this I want it to record if it finds a letter through it. At the end of the loop, I want it to tell me if it found all the letters of the alphabet (a pangram) or if it did not. however I cannot get it to achieve this.

Can someone give me some guidance?

import javax.swing.JOptionPane;
public class CSQ2
{
	public static void main (String [] agrs)
	{
		String sentence = JOptionPane.showInputDialog(null, "Please enter a sentence");
		String result = "";
		String alphabet = "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";
		if (sentence.equals(""))
		result =  "You have not entered a sentence";
		else
		if (sentence.length()<26)
		result = "You have not entered a pangram";
		else
		if (sentence.length()>=26)
		for (int i = 0; i<sentence.length();i++)
		{
			String aChar = sentence.substring(i,i+1);
			String alph = alphabet.substring (i,i+1);
			if(alph.indexOf(aChar)!=-1)
			result = "You have entered a pangram";
			else
			result = "You have entered a pangram";
	}
		JOptionPane.showMessageDialog(null, result);
 
	}
}

Use an array of 26 elements to count the frequency of the characters found in the sentence using PhHein's suggestion of considering the characters as char(s) which are numbers. String methods will help with this. Read the String API to find the methods that will be most helpful. Any of the 26 elements with frequency of zero indicate that the sentence is not a pangram.

Did you review the String API page? Did you happen to notice the charAt() method? After following PhHein's suggestion, it might occur to you that a char value can be translated to the index of a 26-element array, 0 - 25. From there, you can increment the array element based on the char found in the sentence:

// for each character in the sentence, iterated with sentenceIndex:
charToArrayIndex = sentence.charAt( sentenceIndex ) - someOffset;
freqCounterArray[charToArrayIndex]++;

After your update: There's always more than one way. If you like yours and you've tested it and it meets the assignment's requirements, stick with it.

--- Update ---

I can't let it go. Does the code you posted even compile? If so, how did you test it and what results did you get? Did you try testing with "now is the time for all good men to come to the aid of their country"? What sentence did you test with to ensure your code correctly identifies a pangram? How about "abc def ghi jkl mno pqr stu vwx yz"?

Yes it does compile, please feel free to try it yourself.

yes I did test it with "the quick brown fox jumps over a lazy dog" ( for which I got a positive response) and "The quick brown fox jumped over a lazy dog" (for which I got a negative response). I also tried "aacdefghijklmnopqrstuvwxyz" (for which I got a negative response). I did after trying a few other inputs find an error in the code as when I enter "abcdefghijklmnopqrstuvwxyz" I got a negative response but if I enter "abcdefghijklmnopqrstuvwxyz + any other letters") I get a positive reponse. I am trying to figure out logically why this is happening.
I will try your other suggestions now

import javax.swing.JOptionPane;
public class CSQ2
{
	public static void main (String [] agrs)
	{
		String sentence = JOptionPane.showInputDialog(null, "Please enter a sentence");
		String sentenceCopy = sentence;
		sentence = sentence.toLowerCase();
		String result = "";
		String alphabet = "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";
		if (sentence.equals(""))
		result =  "You have not entered a sentence";
		else
		if (sentence.length()<26)
		result = "You have not entered a pangram";
 
		else
		{
			if (sentence.length()>=26)
			for (int i = 0; i<sentence.length();i++)
			{
				String alph = alphabet.substring (i,i+1);
				if(sentence.indexOf(alph)!=-1)
				result = "You have entered a pangram";
				else
				result = "You have entered not entered a pangram";
	}
}
		JOptionPane.showMessageDialog(null, result);
 
	}
}

I'm not sure this approach will ultimately prevail, but I'll play along a bit longer. What are the commas in the String alphabet? Are they necessary? Are they helping? Hurting?

Okay. I was able to spend more time with the code you've posted, and it will work. There are two minor changes to make: the contents of the String 'alphabet' and the boundary of the for loop that checks if the user's sentence is a pangram. You can also do some optimizing as PhHein suggested, but it's fundamentally a good design.

Good job!

One problem I can see is that a StringIndexOutOfBoundsException will occur if the sentence is longer than alphabet.

Without giving away the answer, the problem you've highlighted will be addressed by my suggestion to change "the boundary of the for loop." Almost giving the answer away, the OP misunderstands which length drives the number of passes through the for loop.

If the issue mentioned above is fixed then it will always say it is a pangram if the sentence contains a 'z'. It will only be the last letter of alphabet that sets the final value of result with all others being overwritten.

I don't get this point, but maybe it's because I'm seeing the final solution and you're not.

Also, some of the if statements are pointless. An empty String has a length of 0 which is less than 26 making the check for an empty String redundant. If the String has a length of 26 or greater it must enter the else branch therefore the inner if statement is also redundant.

One was the OP's validity check for user entry, catching the <return> with no characters entered, the other a preliminary check for a pangram since no sentence less than 26 could be one. Is it an accurate validity check? No, but it isn't a flaw in the design that prevents it from identifying pangrams either. As I mentioned, other opportunities to optimize the final solution exist without significantly changing the fundamental approach.

I think it is a bad design.

The approach is sound and can be used to complete a program that satisfies the requirements as described by the OP.

Ahhh, I see. I corrected this logic error in my implementation of the design. Hopefully, the OP's own testing would reveal this flaw.

Guys, thank you very much for your responses.
I have taken the commas out of the String alphabet, I had thought that they were needed in a string, have learned my lesson on this.
I have changed the boundary of the for loop to what I think you were indicating and I have also introduced two counters.
I have tested the code initally and will continue to test it to see if I can find errors.
However one things that is going to happen with this code is regardless of when the program realises it has found that a character of the alphabet does not exist within the String sentence it is going to continue to go through the loop.
Is there a way I can set it to exit the loop as soon as it finds this?
Thanks again for the help, my revised code is below.

public class CSQ2
{
	public static void main (String [] agrs)
	{
		String sentence = JOptionPane.showInputDialog(null, "Please enter a sentence");
		String sentenceCopy = sentence;
		String aLetter = "";
		sentence = sentence.toLowerCase();
		String result = "";
		String alphabet = "abcdefghijklmnopqrstuvwxyz";
		int counterPos = 0;
		int counterNeg = 0;
		if (sentence.equals(""))
		result =  "You have not entered a sentence";
		else
		if (sentence.length()<26)
		result = "You have not entered a pangram";
 
		else
		{
			if (sentence.length()>=26)
			for (int i = 0; i<alphabet.length();i++)
			{
				String alph = alphabet.substring (i,i+1);
				if(alph.indexOf(sentence)!=-1)
				counterPos++;
				else
					{
					if(sentence.indexOf(alph)==-1)
				counterNeg++;
					}			
				}
 
			if(counterNeg>0)
			result = "You have not entered a pangram";
			else
			result = "You have entered a pangram";
		}
		JOptionPane.showMessageDialog(null, result);
 
	}
}

Explore the break; statement.

Remember that you need 26 yay votes to be a pangram but only one nay to know that the user's sentence is not a pangram. You can cease checking as soon as a nay is determined. Review Junky's last post for a good approach.