Simple way to count character occurrences in a string

Question

Is there a simple way (instead of traversing manually all the string, or loop for indexOf) in order to find how many times, a character appears in a string?

Say we have "abdsd3$asda$asasdd$sadas" and we want that $ appears 3 times.

Solution 1

public int countChar(String str, char c)
{
    int count = 0;

    for(int i=0; i < str.length(); i++)
    {    if(str.charAt(i) == c)
            count++;
    }

    return count;
}

This is definitely the fastest way. Regexes are much much slower here, and possible harder to understand.

Solution 2

Functional style (Java 8, just for fun):

str.chars().filter(num -> num == '$').count()

Solution 3

Not optimal, but simple way to count occurrences:

String s = "...";
int counter = s.split("\\$", -1).length - 1;

Note:

• Dollar sign is a special Regular Expression symbol, so it must be escaped with a backslash.
• A backslash is a special symbol for escape characters such as newlines, so it must be escaped with a backslash.
• The second argument of split prevents empty trailing strings from being removed.

Solution 4

You can use Apache Commons' StringUtils.countMatches(String string, String subStringToCount).

Solution 5

Since you're scanning the whole string anyway you can build a full character count and do any number of lookups, all for the same big-Oh cost (n):

public static Map<Character,Integer> getCharFreq(String s) {
  Map<Character,Integer> charFreq = new HashMap<Character,Integer>();
  if (s != null) {
    for (Character c : s.toCharArray()) {
      Integer count = charFreq.get(c);
      int newCount = (count==null ? 1 : count+1);
      charFreq.put(c, newCount);
    }
  }
  return charFreq;
}

// ...
String s = "abdsd3$asda$asasdd$sadas";
Map counts = getCharFreq(s);
counts.get('$'); // => 3
counts.get('a'); // => 7
counts.get('s'); // => 6

View more solutions

Comments

[Mark Amery]

This works, but it's confusing and overcomplicated. Why use recursion - let alone this convoluted implementation - when iteration can be used simply and cleanly? Also, indexOffinds the leftmost index of the target string, so countChars(input.substring(0, input.indexOf(find)), find) will always be equal to zero; you could scrap the first line of your return expression and get the same result.

[jahroy]

Regex = cleaner? Maybe shorter, but I would say regexes are generally pretty cryptic (and less clean) when compared to basic looping code.

[user3090078]

Yahoo! I got voted up! My first post ever :-D

[user3090078]

@Marcelo solution looks nice but fails with: "$$$$$$$$".split("\\$").length - 1 which = -1 or "$$$$$$$$".split("$").length - 1 which = 0.

[noddy]

Almost a good idea. This fails on some simple cases though. e.g. s = "$"

[Dmitry Ginzburg]

Ooh, don't you mind using enhanced for loop instead of C-style going through indices?

[Daniel]

I don't want to use implicit iterators.

[Dmitry Ginzburg]

firstly, surely, there's no shame to use iterators, what's the problem? Secondly, iterators are not used for looping arrays, see, for example, stackoverflow.com/a/7956673/1828937.

[Daniel]

@DmitryGinzburg: The problem is we don't have a char array here, but a String. The alternate way would be to iterate over s.toCharArray(), but this would involve a copy of the Strings internal char array.

[Boann]

Doesn't work because $ is a regex metacharacter.

[Chase]

This is needlessly complex, you can just use an array (256 long if only only basic characters are used, or 65535 long for the entire range java supports). do something like array[str.charAt(i)]++ instead of that if/else branch, and then on the output, just do for(int i = 0; i < length; ++i) if(array[i] != 0) System.out.println((char)i + " = " + array[i]).

[Marcelo]

I corrected the meta character issue in the answer.

[Kaveesh Kanwal]

But, what if we have to find each character's occurrence in the string?

[RajaReddy PolamReddy]

this is the best way :)

[Hirofumi Okino]

That still doesn't work with trailing "$"s, e.g. "abdsd3$asda$asasdd$sadas$" , because empty trailing items are ignored by default. It should be int counter = s.split("\$", -1).length - 1;

[Yeti]

"\$" should be replaced with "\\$", because the backslash already has a special meaning that is unique to the string itself, such as newlines, tabs, etc. (escape character) Therefore a double backslash is needed.

[Sameer Kazi]

waw super like !

[oyenigun]

it doesn't work in case of last char searching ex:"356595".split("5") -> size =3 "356559".split("5") -> size = 4

[Kalpak Gadre]

I always prefer apache commons

[Integraty_beast]

can't understand clearly