Question

class Test{   
    public static void main(String[] args) {     
        String line = "ALONE AMEND DATES DAWNS DEALS LEMON LENDS";
        int maxVowelAtChar = 4;     

        // need the help for this function.  
        String pure = myFunc(line, maxVowelChar);  
        System.out.println(pure);   
    }
} 

Now it should print :

DAWNS LENDS

After maxVowelAtChar there should not be any vowel. the count start from last character of word. “ALONE” does not qualify since it has vowel at char 1 & 3, similarly “LEMON” has vowel at char 2. “DAWNS” qualifies since it has no vowels upto 3rd characters(4th one is A). Help me please.

Answer 1

EDITED: This code passes the clarified requirements. Changed to count from end backwards.

Before you criticise, see notes below.

public static void main(String[] args)
{
    String line = "ALONE AMEND DATES DAWNS DEALS LEMON LENDS";
    for (int maxVowelAtChar = 1; maxVowelAtChar < 6; maxVowelAtChar++)
    {
        String pure = myFunc(line, maxVowelAtChar);
        System.out.println(maxVowelAtChar + ": " + pure);
    }
}

private static String myFunc(String line, int maxVowelAtChar)
{
    String pure = "";

    for (String word : line.split("\\s"))
    {
        // This regex is also sensitive to word length ie not being 5 like all test data
        if (word.matches("^.{" + (word.length() - maxVowelAtChar + 1) + "}[^aeiouAEIOU]*$"))
            pure += word + " ";
    }

    return pure.trim();
}

Output:
1: ALONE AMEND DATES DAWNS DEALS LEMON LENDS
2: AMEND DATES DAWNS DEALS LEMON LENDS
3: AMEND DAWNS DEALS LENDS  
4: DAWNS LENDS
5: 

This code has been simplified to illustrate the core of the solution. “Real” code would use a StringBuffer instead of String += String.

What makes this work is the dynamic regex passed to matches that means “must end with n non-vowels”