Thursday, July 30, 2015



TopCoder Problem Statement - WordForm
Single Round Match 173 Round 1 - Division I, Level One
Single Round Match 173 Round 1 - Division II, Level Two


Detect vowels and consonants in a word.

Java Source:
     1 public class WordForm {
     3  public String getSequence(String word) {
     5         char[] cvArray = new char[word.length()];
     7         boolean prevIsVowel = true;
     9         for (int i = 0; i < word.length(); i++)  {
    10             cvArray[i] = (prevIsVowel = isVowel(word.charAt(i), prevIsVowel)) ? 'V' : 'C';
    11         }
    13         // Remove duplicate characters
    14         return new String(cvArray).replaceAll("([CV])\\1+", "$1");
    15  }
    17     private static boolean isVowel(Character c, boolean prevIsVowel)  {
    19         c = Character.toUpperCase(c);
    21         if ((c.equals('A')) || (c.equals('E')) ||
    22                 (c.equals('I')) || (c.equals('O')) ||
    23                 (c.equals('U'))) return true;
    25         if (!c.equals('Y')) return false;
    27         return !prevIsVowel;
    28     }
    29 }

The problem can be solved by breaking it up into two pieces. First, obviously we're going to need a method to determine if a character is a consonant or vowel. This is handled by isVowel(). isVowel() first converts the character to upper-case, and then compares it to the standard vowels A, E, I, O, and U. If it matches one of these, or if it doesn't and also is not a 'Y' then we're done. 'Y's are the only tricky letter.

To determine if 'Y' is a vowel or not, isVowel() requires a boolean denoting whether the previous character was a vowel or not. We can set prevIsVowel's initial value to true since the behavior of a 'Y' at the beginning of a string is the same as a 'Y' following a vowel.

With a reliable method of determining consonants and vowels, we just need to construct the return String, and the only challenge there is eliminating duplicates. One way to do this would be to keep track of the last character, and only append the next character if it differed. An easier way is to just add all the characters, and then use a regular expression to remove the duplicates. This is done in the replaceAll() method of the return statement.

If you're not familiar with the assignment on line 10, all this does is assign the result of isVowel() to both prevIsVowel and cvArray[i] in one line. First, isVowel() is evaluated and the result is assigned to prevIsVowel. The result of this assignment, which is either true or false, is then assigned to cvArray[i]. With that, we've put the correct character (either a 'V' or a 'C') into the array, and set prevIsVowel to the correct value for the next call into isVowel().

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