Count the number of times a character appears in a contiguous manner in a string

Question

I'm new to Java. I'm trying to print the characters present in the string along with their count. The count only increments when the same character is present next to it.

Ex:

I/O : Sssgs

O/P : S1s2g1s1

Counting the occurence of each character gives the count of the full count regardless of the characters not being present next to each other. Tampering with the i & j loops gives an OutOfBounds error.

      //ch[] is the String converted to a character array.
     //count[] is an array to store count of the characters      

    //Checks if present char and next char are same and increments count
    for(int i=0;i<ch.length;i++)    
    {
        count[i]=0;
        for(int j=0;j<ch.length;j++)
        {
            if(ch[i]==ch[j])
            {
                count[i]++;
            }
        }
    }

    //Prints Distinct char
    for(int i=0;i<ch.length;i++)
    {
        int j;
        for(j=0;j<i;j++)
        {
            if(ch[i]==ch[j])
            {
                break;
            }
        }

        if(i==j)
        {
            System.out.print(ch[i]+" "+count[i]);
        }
    }

The Input is > HelloWorld

The expected output should be > H1 e1 l2 o1 W1 o1 r1 l1 d1

I would use a `StringBuiler`. 1) iterate over the string, 2) add the character to the builder and increment the counter 3.a) if previous character is equal to the current, increment the counter, 3.b) if it is not equal, append the counter and reset it -> continue at 2. — Lino, Apr 02 '19 at 11:25
Start anew, and "program" with paper & pen. One only needs to know the prior character (char) and how many times it occured (int), is it not? Or the current character (char) and how many times it follows (int). The solution is short. — Joop Eggen, Apr 02 '19 at 11:28
Do you mean 'contiguous' (next to one another)? 'Contagious' means likely to spread, as in like a disease. — Michael, Apr 02 '19 at 11:30
Please read "How to create a [mcve]". Then use the [edit] link to improve your question (do not add more information via comments). Otherwise we are not able to answer your question and help you. You give the expected output, omitting the actual output. And when you have exceptions, include the exception stack trace. — GhostCat, Apr 02 '19 at 11:56
Beyond that, have a look at: https://stackoverflow.com/questions/5554734/what-causes-a-java-lang-arrayindexoutofboundsexception-and-how-do-i-prevent-it — GhostCat, Apr 02 '19 at 11:57

score 1 · Accepted Answer · answered Apr 02 '19 at 11:45

I just made some corrections in your code and below it is how it looks like:

public static void main(String[] args) {
    String s = "Sssgs";
    char[] ch = s.toCharArray();
    int[] count = new int[20];

       for(int i=0;i<ch.length;i++)    
        {
            count[i]=0;
            for(int j=i;j<ch.length;j++)
            {
                if(ch[i]==ch[j])
                {
                    count[i]++;
                } else {
                    break;
                }
            }
        }

        //Prints Distinct char
        for(int i=0;i<ch.length;i += count[i])
        {
            System.out.print(ch[i] + "" +count[i]);
        }
}

Most changes was in Prints Distincts when I just read character and it number of occurence and then jump that number in iteration. It lets me stops on the next character which is different

The output for "Sssgs" is "S1s2g1s1" and for "HelloWorld" is "H1e1l2o1W1o1r1l1d1"

This solved my problem, thanks. I am confused on the Distinct's for loop. What exactly does `i += count[i]` do? — Misr, Apr 02 '19 at 12:35
It checks the number of occurence of letter and jump that number. So if you have "sssS" so for i=0 ('s' character) count = 3 because there are triple 's' so the next 'i' will be 0 + 3 = 3 so it jumps to the letter 'S' and check it's occurence which is 1 then i = 3 + 1 = 4 which is not less than ch.length so the loop is finished — Rafał Sokalski, Apr 02 '19 at 12:41

score 1 · Answer 2 · answered Apr 02 '19 at 12:08

1

Here is a simpe solution that doesn't use any extra array and instead directly prints the counted char when the next one is different

char prevChar = ch[0];
int count = 1;
for (int i = 1; i < ch.length; i++) {
  if (ch[i] != prevChar) {
    System.out.printf("%c%d ", prevChar, count);
    count = 1;
    prevChar = ch[i];
  } else {
    count++;
  }
}
System.out.printf("%c%d ", prevChar, count);

answered Apr 02 '19 at 12:08

Joakim Danielson

43,251
5
22
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This looks like the perfect solution to me and should be the accepted answer. Alternatively the result could be written into a StringBuilder or an ArrayList. – Amadán Apr 02 '19 at 13:18

user2087103 · Answer 3 · 2019-04-02T12:03:04.137

I hate this solution, but I guess you are using a char[] because of needs. If not compulsory, i would recommend you to use a StringBuilder as Lino suggested.

char blankChar = " ".charAt(0);
if (stringInput == null || "".equals(stringInput)) {
    System.out.println("Empty input");
}
char[] ch = stringInput.toCharArray();
char lastChar = ch[0];
int numAppearanceslastChar = 0;
for (char element : ch) {
    if (element == blankChar) {
        continue;
    }
    if (lastChar == element) {
        numAppearanceslastChar++;
    } else {
        System.out.print(lastChar+""+numAppearanceslastChar+" ");
        lastChar = element;
        numAppearanceslastChar = 1;
    }
}
System.out.println(lastChar+""+numAppearanceslastChar+" ");

Output: H1 e1 l2 o1 w1 o1 r1 l1 d1

Explanation: Read the whole word only once (note you wuere doing 3 times a for loop) and compare last read char with the new one. If they match, increment the number of occurences of that char. If they are not, then you print them and set the new characted as the last read. When you end reading the word, print the last char read.

Always remember to keep it simple! And sanitize (you would get a nullPointer in this code if receiving a null or empty, just wrote it there to point it out).

Why exactly are you introducing a String, just to get the char[] out of it? Also, what is the idea behind char blankChar = " ".charAt(0) - instead of just doing char blankChar = ' '? — Amadán, Apr 02 '19 at 13:23
I was using a String just because I tested it with a String, not a char[], and Im more used to using Strings. Indeed, I would do something like Lino's, but worked it out using a char[]. The idea between using charAt(0) and not '' is the same one, being used to work with Strings and not chars. :) — user2087103, Apr 04 '19 at 11:02

score 0 · Answer 4 · answered Apr 02 '19 at 12:01

I came up with this:

public static String count(String in) {
    if (in == null || in.isEmpty()) {
        return in;
    }
    int length = in.length();
    if (length == 1) {
        return in + '1';
    }
    StringBuilder out = new StringBuilder(length << 1);

    char previous = in.charAt(0);
    int count = 1;
    for (int i = 1; i < length; i++) {
        char current = in.charAt(i);
        if (previous == current) {
            count++;
        } else {
            out.append(previous).append(count);
            previous = current;
            count = 1;
        }
    }
    return out.append(previous).append(count).toString();
}

Takes care of null and empty strings. And strings with length == 1 (which is just string + 1).

This solution also doesn't need to create an additional char[] array, as it is working with charAt

Count the number of times a character appears in a contiguous manner in a string

4 Answers4