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;
}

Hint: you want to check OR conditions with the `||` operator. — moonwave99, Dec 10 '15 at 18:31
Can you please add the current output of your program and the desired one? Consider adding a description about the difference. — Marcinek, Dec 10 '15 at 18:39
Possible duplicate of [Java - Writing a syllable counter based on specifications](http://stackoverflow.com/questions/9154027/java-writing-a-syllable-counter-based-on-specifications) — Frakcool, Dec 10 '15 at 18:43

Ankit Deshpande · Accepted Answer · 2015-12-11T17:48:38.330

  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.

Thanks for your quick reply. Your answer worked great however if there was a string such as 'be' or 'mmmmeeeee' or 'the', countSyllables would return 0 except it should be 1. I'm not sure how I would include this. — Don Lim, Dec 11 '15 at 05:30
Can you please clarify this? According to the problem statement answer for 'obstinanceeeeeee' is 3 and why is it 1 for 'be'? How do you want to treat the 'e's in the end of the String or how do you differentiate it? is it that if there's a single 'e' in the end, it has to counted? — Ankit Deshpande, Dec 11 '15 at 12:25
Yes. For words such as 'me', 'the', and 'be', while (i >= 0 && input.charAt(i) == 'e') { i--; } gets rid of the character 'e' but will return syllables = 0. — Don Lim, Dec 11 '15 at 16:49
I have made corresponding changes. It will work now. Check the edited ans — Ankit Deshpande, Dec 11 '15 at 17:41
Thank you again for your help. I added 2 sections of code that would return 1 syllable if there is a string such as 'sdlfjse' (any string with all consonants and an 'e' at the end) but not double count syllables for strings such as 'one'. — Don Lim, Dec 11 '15 at 18:49

Counting syllables within a word

1 Answers1