check number of a given character occurence in a String

Question

I want to write a simple class to process Strings (may be very long strings up to 1mil characters in it). String will basically consists of two characters \"a\" and \"b\" that

Answer 1

What's wrong with using something simple as this? Your idea of doing something as simple as this is an overkill, and would end up using more resources.

String s = "abbb";

int a = 0;
int b = 0;
for(int i = 0; i<s.length(); i++){
    if((s.charAt(i) == 'a')){
        a += 1;
    } else {
        b += 1;
    }
}

a = 1; b = 3

Answer 2

If you want you can use a third party library like StringUtils. It has a method countMatches which will do the work.

StringUtils.countMatches("abba", "a")   = 2
StringUtils.countMatches("abba", "ab")  = 1

Answer 3

I believe this is what you want:

private static boolean check(String input) {
    int count = 0;
    for (int i = 0; i < input.length(); ++i) {
        if (input.charAt(i) == 'a') {
            count++;
        }
    }
    return count == input.length() >> 1; // count == input.length()/2
}

Answer 4

Why do you need regular expression and split the string for this! You can simply loop through the string and count the number of a and bs. You need to keep two different counter, one for a and one for b. Using regular expression will be less efficient. There is no way you can get the result without traversing the string at least once. So use a simple loop to count a and b.

• You can make one optimization in the loop. If anytime mod of countA - countB is greater than the number of remaining characters then a and b can never be equal. So you can break the loop then.

• If the length of the string is odd then there is no need to count. Count of a and b can never be equal when total number of elements is odd.

Answer 5

public class Typo { 
    public static void main(String[] args){     
        String ver = "NOK";

        String text = "ababababbaba";

        if( (text.length() - text.replaceAll("a","").length() ) ==  
            ( text.length() - text.replaceAll("b","").length() ) ) {
            ver = "OK";     
        }

        System.out.println(ver);
    }
}

Answer 6

You should definitely not using regexp for this problem: generally speaking, regexp is not good when you need to count anything. You cannot even write a regexp to check if brackets in an expression are balanced.

For this problem a simple counter will be sufficient: increment on a, decrement on b, check for zero in the end to know the answer to your problem.

boolean check(String s) {
    int count = 0;
    for (int i = 0 ; i != s.length() ; i++) {
        if (s.charAt(i) == 'a') {
            count++;
        } else { /* it is b */
            count--;
        }
    }
    return count == 0;
}