WordDistanceFinder solution

WordDistanceFinder.java

/* This class will be given a list of words (such as might be tokenized
 * from a paragraph of text), and will provide a method that takes two
 * words and returns the shortest distance (in words) between those two
 * words in the provided text.
 * Example:
 *   WordDistanceFinder finder = new WordDistanceFinder(Arrays.asList("the", "quick", "brown", "fox", "quick"));
 *   assert(finder.distance("fox", "the") == 3);
 *   assert(finder.distance("quick", "fox") == 1);
 *
 * "quick" appears twice in the input. There are two possible distance values for "quick" and "fox":
 *     (3 - 1) = 2 and (4 - 3) = 1.
 * Since we have to return the shortest distance between the two words we return 1.
 */

 class WordDistanceFinder {
   final List<String> words;

    public WordDistanceFinder (List<String> words) {
      this.words = words;
    }
    
    public int distance (String wordOne, String wordTwo) {
      if (wordTwo == null || wordOne == null) {
        throw new NullPointerException("Words cannot be null");
      }
      
      if (wordOne.equals(wordTwo)) {
        return 0;
      }
      
      Integer lastWordOneIndex = null;
      Integer lastWordTwoIndex = null;
      Integer minValue = Integer.MAX_VALUE;
      for (int i = 0; i < words.size(); i++) {
        final String currentWord = words.get(i);
        if (wordOne.equals(currentWord)) {
          lastWordOneIndex = i;
          if (lastWordTwoIndex != null) {
            minValue = Math.min(minValue, lastWordOneIndex - lastWordTwoIndex);
          }
        } else if (wordTwo.equals(currentWord)) {
          lastWordTwoIndex = i;
          if (lastWordOneIndex != null) {
            minValue = Math.min(minValue, lastWordTwoIndex - lastWordOneIndex);
          }
        }
      }
      
      if (lastWordTwoIndex == null || lastWordOneIndex == null) {
        throw new InvalidParameterException("Words are not on the list");
      }
      
      return minValue;
    }
  }