244. Shortest Word Distance II

Problem:

This is a follow up of Shortest Word Distance. The only difference is now you are given the list of words and your method will be called repeatedly many times with different parameters. How would you optimize it?

Design a class which receives a list of words in the constructor, and implements a method that takes two words word1 and word2 and return the shortest distance between these two words in the list.

For example,
Assume that words = ["practice", "makes", "perfect", "coding", "makes"].

Given word1 = “coding”, word2 = “practice”, return 3.
Given word1 = "makes", word2 = "coding", return 1.

Analysis:

pre process array, store each words occurrence index to hashtable. Then compare distance of two words' indexes. Suppose we have word1 with indexes [2,5,8,9], word2 with indexes [4,7,11]. We can compare the starting indexces first, then increment the smaller one by one to make them closer until we reach one of the indexes array's end. 

Solution:

class WordDistance {
    Map<String, List<Integer>> map;
    public WordDistance(String[] words) {
        map = new HashMap<>();
        for(int i = 0; i < words.length; i++) {
            map.computeIfAbsent(words[i], k -> new ArrayList<>()).add(i);
        }
    }
    
    public int shortest(String word1, String word2) {
        List<Integer> list1 = map.get(word1);
        List<Integer> list2 = map.get(word2);
        int res = Integer.MAX_VALUE;
        int i = 0, j = 0;
        while (i < list1.size() && j < list2.size()) {
            res = Math.min(Math.abs(list1.get(i) - list2.get(j)), res);
            if (list1.get(i) < list2.get(j)) {
                i++;
            } else {
                j++;
            }
        }
        return res;
    }

}