Counting syllables within a word

Question

I am looking to create a method that counts the number of syllables within a word. A syllable defined in my project as a contiguous group of vowels but not an 'e' at the end of the word. So the word 'obvious' by this definition only has 2 syllables and the string 'obstinanceeeeeee' has 3 syllables. My attempt is:

protected int countSyllables(String word)
String input = word.toLowerCase();
    int i = input.length() - 1;
    int syllables = 0, numOfE = 0;
    // skip all the e's in the end
    while (i >= 0 && input.charAt(i) == 'e') {
        i--;
        numOfE++;
    }

    // This counts the number of consonants within a word
    int j = 0;
    int consonants = 0;
    while (j < input.length()) {
        if (!isVowel(input.charAt(j))) {
            consonants++;
        }
        j++;
    }

    // This will return syllables = 1 if the string is all consonants except for 1 e at the end.
    if (consonants == input.length() - 1 && numOfE == 1) {
        syllables = 1;
    }

    boolean preVowel = false;
    while (i >= 0) {
        if (isVowel(input.charAt(i))) {
            if (!preVowel) {
               syllables++;
               preVowel = true;
            }
        } else {
            preVowel = false;
        }
        i--;
    }
    return syllables;
}

public boolean isVowel(char ch) {
   if (ch == 'a' || ch == 'e' || ch == 'i' || ch == 'o' || ch == 'u' || ch == 'y') {
       return true;
   }
   return false;
}

Answer 1

  protected int countSyllables(String word) {
    String input = word.toLowerCase();
    int i = input.length() - 1;
    // skip all the e's in the end
    while (i >= 0 && input.charAt(i) == 'e') {
        i--;
    }
    int syllables = 0;
    boolean preVowel = false;
    while (i >= 0) {
        if (isVowel(input.charAt(i))) {
            if (!preVowel) {
               syllables++;
               preVowel = true;
            }
        } else {
            preVowel = false;
        }
        i--;
    }
        return syllables;
    }

    public boolean isVowel(char ch) {
       if (ch == 'a' || ch == 'e' || ch == 'i' || ch == 'o' || ch == 'u' || ch == 'y') {
           return true;
       }
       return false;
    }

I hope this helps. You can start iterating from the end of the string and ignore all the e's before processing/counting the syllables.
Making edit to the original answer. Now will count the word as syllable if it has only one e in the end

 protected int countSyllables(String word) {
    String input = word.toLowerCase();
    int syllables = 0,numOfEInTheEnd=0;

    int i = input.length() - 1;
    // count all the e's in the end
    while (i >= 0 && input.charAt(i) == 'e') {
        i--;
        numOfEInTheEnd++;
    }

    if (numOfEInTheEnd == 1) {
        syllables = 1;
    }

    boolean preVowel = false;
    while (i >= 0) {
        if (isVowel(input.charAt(i))) {
            if (!preVowel) {
               syllables++;
               preVowel = true;
            }
        } else {
            preVowel = false;
        }
        i--;
    }
    return syllables;
    }

    public boolean isVowel(char ch) {
    if (ch == 'a' || ch == 'e' || ch == 'i' || ch == 'o' || ch == 'u' || ch == 'y') {
        return true;
    }
    return false;
    }

Instead of skipping all the trailing e's, now its counting it. If trailing e's is equal to 1, syllables is set to 1. This will work now.