how to count many times a character occurs in a string without using s loop

Question

the code below is meant to count each time character 'x' occurs in a string but it only counts once ..

I do not want to use a loop.

public class recursionJava

{

   public static void main(String args[])

   {

      String names = "xxhixx";

      int result = number(names);
      System.out.println("number of x: " + result);
   }

   public static int number (String name)
   {

      int index = 0, result = 0;

      if(name.charAt(index) == 'x')
      {
         result++;
      }
      else
      {
         result = result;
      }
            index++;

      if (name.trim().length() != 0)
      {
        number(name);
      }
      return result;
   }
}

"I do not want to use a loop" = "I must use recursion because the homework says so" — Kayaman, Mar 05 '18 at 08:39
Possible duplicate of [Java: How do I count the number of occurrences of a char in a String?](https://stackoverflow.com/questions/275944/java-how-do-i-count-the-number-of-occurrences-of-a-char-in-a-string) — Oleksandr Pyrohov, Mar 05 '18 at 13:50

Tim Biegeleisen · Answer 1 · 2018-03-05T08:38:20.863

6

You could do a replacement/removal of the character and then compare the length of the resulting string:

String names = "xxhixx";
int numX = names.length() - names.replace("x", "").length(); // numX == 4

edited Mar 05 '18 at 08:38

answered Mar 05 '18 at 08:37

Tim Biegeleisen

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Is this really more efficient (and idiomatic) than cycling through the string? – Federico klez Culloca Mar 05 '18 at 08:37
1

@FedericoklezCulloca Because it answers the question, – Scary Wombat Mar 05 '18 at 08:39
@FedericoklezCulloca It is common knowledge that regex type solutions, while elegant, can perform poorly as compared to direct base string operations. So this may not be the most performant solution. – Tim Biegeleisen Mar 05 '18 at 08:39
@FedericoklezCulloca It's obvious using simple loop gives the best performance, and OP requests not to do this. – Jai Mar 05 '18 at 08:41
In a similar vain: `names.replaceAll("[^x]", "").length()`. Alternatively, `names.split("x", -1).length()-1`. – Andy Turner Mar 05 '18 at 08:52

Eran · Answer 2 · 2018-03-05T08:46:53.670

5

If you don't want to use a loop, you can use recursion:

public static int number (String name)
{
    if (name.length () == 0)
        return 0;
    int count = name.charAt(0)=='x' ? 1 : 0;
    return count + number(name.substring(1));
}

edited Mar 05 '18 at 08:46

answered Mar 05 '18 at 08:37

Eran

387,369
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Interesting that people aren't piling on this question complaining about its (lack of) efficiency ([vs this](https://stackoverflow.com/a/49106335/3788176)). – Andy Turner Mar 05 '18 at 08:46
I didn't say it was: I'm just pointing out that there is a chain of comments on the other answer complaining that isn't efficient, but not here. – Andy Turner Mar 05 '18 at 08:48
I'm not criticizing either answer. – Andy Turner Mar 05 '18 at 08:50

score 4 · Answer 3 · answered Mar 05 '18 at 08:45

4

As of Java 8 you can use streams:

"xxhixx".chars().filter(c -> ((char)c)=='x').count()

answered Mar 05 '18 at 08:45

Luk

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score 4 · Answer 4 · answered Mar 05 '18 at 08:56

Previous recursive answer (from Eran) is correct, although it has quadratic complexity in new java versions (substring copies string internally). It can be linear one:

    public static int number(String names, int position) {
    if (position >= names.length()) {
        return 0;
    }

    int count = number(names, position + 1);

    if ('x' == names.charAt(position)) {
        count++;
    }
    return count;
}

hensing1 · Answer 5 · 2018-03-05T10:38:52.297

Your code does not work because of two things:

Every time you're calling your recursive method number(), you're setting your variables index and result back to zero. So, the program will always be stuck on the first letter and also reset the record of the number of x's it has found so far.
Also, name.trim() is pretty much useless here, because this method only removes whitespace characters such as space, tab etc.

You can solve both of these problems by

making index and result global variables and
using index to check whether or not you have reached the end of the String.

So in the end, a slightly modified (and working) Version of your code would look like this:

public class recursionJava {

    private static int index = 0;
    private static int result = 0;

    public static void main(String[] args) {
        String names = "xxhixx";

        int result = number(names);
        System.out.println("number of x: " + result);
    }

    public static int number (String name){
        if(name.charAt(index) == 'x')
            result++;

        index++;

        if(name.length() - index > 0)
            number(name);
        return result;
    }

}

score 1 · Answer 6 · answered Mar 05 '18 at 08:42

1

You can use StringUtils.countMatches

StringUtils.countMatches(name, "x");

answered Mar 05 '18 at 08:42

Pelin

936
5
12

That uses a loop internally. – Kayaman Mar 05 '18 at 08:44
2

@Kayaman So does the String method `replace` used in the current highest voted answer. – OH GOD SPIDERS Mar 05 '18 at 08:54
@OHGODSPIDERS that's true, but you're not really putting them on the same line are you. – Kayaman Mar 05 '18 at 08:57
I think recursion is an interesting solution and I would recommend to use this approach during a job interview etc. , but for a real life project the StringUtils class provides simpler and cleaner usage. – Pelin Mar 05 '18 at 09:30
I suppose you are referring to [Appache Commons `StringUtils.countMatches`](https://commons.apache.org/proper/commons-lang/apidocs/org/apache/commons/lang3/StringUtils.html#countMatches-java.lang.CharSequence-char-) - might be worth mentioning that. – Hulk Mar 05 '18 at 12:05

how to count many times a character occurs in a string without using s loop

6 Answers6