Count using Enhanced for loop

Question

I have this code below, my goal is to count how many letter 'e' is in the String "abcdee".

class Sample1 {
    String tiles;

    public Sample1 (String tiles) {
        this.tiles = tiles;
    }

    public int countLetter(char letter) {
        int a = 0;      
        for (char x : tiles.toCharArray()) {
            int m = 0;
            if (tiles.indexOf(letter) != -1) {
                m = 1;
            }
            a += m;
            System.out.println("Letter count is " + m); 
        }
        return a;
    }
}

public class Sample {

    public static void main(String[] args) {
        Sample1 s = new Sample1("abcdee");
        s.countLetter('e'); 

    }
}

I expect that the code would give me this result:

Letter count is 0
Letter count is 0
Letter count is 0
Letter count is 0
Letter count is 1
Letter count is 1

and then maybe add all the 1's to get 2. But all I get is this when I run it:

Letter count is 1
Letter count is 1
Letter count is 1
Letter count is 1
Letter count is 1
Letter count is 1

Hope you can help me out.

Are you meaning to print out the value of a instead of the value of m? — Nick Acosta
– Nick Acosta, Commented May 28, 2017 at 0:34
One-liner using streams: return tiles.chars().filter(c -> c == letter).count(). — beatngu13
– beatngu13, Commented May 28, 2017 at 0:47
@beat did you mean Arrays.toStream(tiles.toCharArray()).filter(c -> c == letter).count() — Jigar Joshi
– Jigar Joshi, Commented May 28, 2017 at 1:22
tiles.indexOf(letter) will output valid non negative index if letter is present in titles. in your code both are not changing during the loop execution. abcdee does contain e hence its printing like that — nits.kk
– nits.kk, Commented May 28, 2017 at 2:17
@JigarJoshi no, I mean CharSequence#chars(), which directly gives you a stream of characters. — beatngu13
– beatngu13, Commented May 28, 2017 at 8:40

halfer · Accepted Answer · 2019-09-21 22:25:24Z

3

The easiest way to fix your code is to change the logic in your counting method to

    int a = 0;      
    for (char x : tiles.toCharArray()) {
        if (x == letter) {
            a += 1;
        }
    }
    return a;

There are better ways, though.

You might want to look as this old Stack Overflow question which is a more general solution to the problem you are working on.

edited Sep 21, 2019 at 22:25

halfer

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answered May 28, 2017 at 0:35

Ray Toal

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1 Comment

Arcee Over a year ago

Thanks Ray I should have kept it simple... thanks for reminding me...you're a ninja.. :)

Snehal · Accepted Answer · 2017-05-28 01:18:48Z

1

The indexOf(String target) method searches left-to-right inside the given string for a "target" string. The indexOf() method returns the index number where the target string is first found or -1 if the target is not found. So if 'e' is present it would return true . You have not used variable x anywhere in your loop first of all .Maybe you can try ,

if (x == letter) {
  a+ = 1
}

instead of

if (tiles.indexOf(letter) != -1) {
  m = 1;
}

answered May 28, 2017 at 1:18

Snehal

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Comments

Ousmane D. · Accepted Answer · 2017-05-28 00:58:24Z

0

It's repeatedly printing 1 because you're using indexOf method, that means as long as the letter e exists within the string tiles it will always print Letter count is 1.To solve the problem at hand simply change:

if (tiles.indexOf(letter) != -1)

to:

if (x == letter)

An even simpler solution to return the letter count as of java-8 is:

public int countLetter(char letter) {
       return (int)tiles.chars().filter(e -> e == letter).count();
}

edited May 28, 2017 at 0:58

answered May 28, 2017 at 0:43

Ousmane D.

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1 Comment

beatngu13 Over a year ago

Why don't you use chars() when you apply a character-based filter anyway?

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