regionMatches() is a method on String, not on arrays.

   If you're using 1.5, declare your vectors as containg strings.
Otherwise, extract the value from the vector and cast it to string.

   You're gonna have problems if your input data was such that you needed
to remove two consecutive items from myVector2.

   - Oliver

nemadrias schrieb:

I consider that there are at least two entries where the three letters
at the end differ.

If you put Strings into a Collection (Vector implements this interface),
you can use String#regionMatches on it's elements, of course.

If you use Java >= 1.5 you can use generics to make the Vector type-safe:

Vector<String> myVector = new Vector<String>();

Then you can use regionMatches without a cast:
  if ( myVector.get(i).regionMatches(...

If you don't use Java >= 1.5 you have to cast:

  if ( ((String)myVector.get(i)).regionMatches(...

BTW: if you don't access myVector in more than one thread, you should
use ArrayList instead of Vector since Vector is synchronized.

Second: you could have used a custom class to represent your entries.

class Entry {
    private String content;

    public Entry( String content ) {
        if ( content == null )
            throw new IllegalArgumentException(
                    "Cannot have null content.");

        if ( content.length() < 15 )
            throw new IllegalArgumentException(
                    "Content too short" );

        this.content = content;
    }

    public String toString() {
        return content;
    }

    public boolean equals( Object o ) {
        if ( o == null || !(o instanceof Entry) )
            return false;
        return (content.regionMatches(0, o.toString(), 0, 15));
    }

    public int hashCode() {
        return content.substring(0, 15).hashCode();
    }
}

With this class, you can manage your entries in the form of collections
(Java 1.5 syntax):

HashSet<Entry> firstSet = new HashSet<Entry>();
HashSet<Entry> secondSet = new HashSet<Entry>();

// fill the sets, e. g. at loading time or if you only have
// vectors:
// for ( String s : myVector )
//    firstSet.add( new Entry(s) );
// for ( String s : myVector2 )
//    secondSet.add( new Entry(s) );

// now, remove the duplicates:
firstSet.removeAll(secondSet);

Of course, this works with any Collection not just sets.

Third: ... there are many ways...

Bye
Michael

why don't you stick the number in a hashmap, then before you add the next
number, check if it is in the hashmap, if it is not add it.
And if it is already in the map , then do whatever you are going to do.

I think you will find it faster than double looping

steve