不在字符串数组中的字符串中的单词数

Question

I want to create a method that returns the number of words in a string that have no occurrences of words in the array of strings. I want to implement this logic only using anything in the java.lang package.

public int count(String a, String[] b) {

}

Eg

count("  hey   are  you there    ", new String[]{ "are", "i", "am"})

would return 3 as there is the word "are" in the string.

First off, I think I have to use the string.split function to convert the string to an array of strings. Any ideas?

Answer 1

You could simply do something like:

public int count(String a, String[] b) {
    int count = b.length;
    for(String s : b) if(a.contains(s)) count--;
    return count;
}

EDIT: I might have been confused, I thought you wanted the # of strings in b not in a (in your example it would still be 3). In that case, from your example, split seems inconvenient unless you use regex , so you could create a String[] using Scanner :

public int count(String a, String[] b) {
    ArrayList<String> words = new ArrayList<String>();
    Scanner scan = new Scanner(a);
    while(scan.hasNext()) words.add(scan.next());

    int count = words.size();
    for(String s : words) if(/*b contains s*/) count--;
    return count;
}

Answer 2

You logic should go somewhat like this:

1. Split a , right. Now you have a list of words. In a real life, you should probably also try to clarify the requirement—what exactly is a “word”? A reasonable assumption is that it's a sequence of non-whitespace characters, but could be something different (for example, a sequence of letters).

2. Iterate over a and check whether each word is in b . If it isn't, increment your counter. But every check is a linear search in b , leading to the total complexity of O(nm), so...

3. Before iterating, convert b into a HashSet . This is a linear operation, but then your main loop will also become a linear operation, therefore the total complexity will be O(m + n).

4. If you have to do this thing repeatedly for different strings, but the same word list, consider creating a WordCounter class so you only have to create the HashSet once in the constructor.

Answer 3

Follow the steps to complete the task.

• Use StringTokenizer to tokenize the String a .
• Convert String Array b to Collection , so that you can check if it contains the given token.
• Use loop to get next token from StringTokenizer and check if it contains in List .

-

Try below code, it'll work.

EDIT : Using java.util package.

public int count(String a, String[] b) {
    java.util.StringTokenizer tokenizer = new java.util.StringTokenizer(a);
    java.util.List bList = java.util.Arrays.asList(b);
    int tokens = tokenizer.countTokens();
    int counter = tokens;
    for(int i=0;i<tokens;i++) {
        String token = tokenizer.nextToken().trim();
        if(bList.contains(token)) {
            counter--;
        }
    }
    return counter;
}

By using this, you can get the counter in just one for loop.

EDIT :: Using java.lang package only.

public int count(String a, String[] b) {
    String[] words = a.split(" ");
    int tokens = words.length;
    int wordCount = 0;
    int counter = 0;
    for(int i=0;i<tokens;i++) {
        String token = words[i].trim();
        if(token.length() <= 0) {
            continue;
        }
        wordCount++;
        for(String bItem : b) {
            if(bItem.equals(token)) {
                counter++;
                break;
            }
        }
    }
    return wordCount - counter;
}