Codingbat countYZ

user_16561343 0
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Task from Codingbat :

Given a string, count the number of words ending in 'y' or 'z' -- so the 'y' in "heavy" and the 'z' in "fez" count, but not the 'y' in "yellow" (not case sensitive). We'll say that a y or z is at the end of a word if there is not an alphabetic letter immediately following it. (Note: Character.isLetter(char) tests if a char is an alphabetic letter.)

countYZ("fez day") → 2
countYZ("day fez") → 2
countYZ("day fyyyz") → 2

I'm trying to solve this task like this:

public int countYZ(String str) {
      String regex = "(.[.^y^z]\\b)";
      return str.toLowerCase().replaceAll(regex, "").length();
    }

But not all the tests were passed. How can I fix regex = "(.[.^y^z]\b)" in order to pass all the tests?

regex java
2 Answers

You could capture either y or z in a capture group not followed by a char a-z, and match any other char.

In the replacement use capture group 1.

([yYzZ](?![A-Za-z]))|.

Explanation

  • ( Capture group 1
    • [yYzZ] Match y or z in lower or uppercase
    • (?![A-Za-z]) Negative lookahead, assert not a char A-Za-z directly to the right
  • ) Close group 1
  • | Or
  • . Match any other character

See a regex demo and a Java demo.

Example

public int countYZ(String str) {
  String regex = "([yYzZ](?![A-Za-z]))|.";
    return str.toLowerCase().replaceAll(regex, "$1").length();
}

Or the unicode variant for a broader match:

public int countYZ(String str) {
  String regex = "([YyZz])(?!\\p{IsAlphabetic})|.";
    return str.toLowerCase().replaceAll(regex, "$1").length();
}
user_54249880

Better to restrict using regex where you can do so. You can make the solution like this:

public static int countYZ(String str) {
    String[] strings = str.toLowerCase().split("\\s");
    int n = 0;
    for (String s : strings) {
        if (s.endsWith("y") || s.endsWith("z")) {
            n++;
        }
    }
    return n;
}
user_191295320
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