244 Shortest Word Distance II

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.

Note: You may assume that word1 does not equal to word2, and word1 and word2 are both in the list.

The Idea: First I considered doing O(n^2) preprocessing and create a lookup table to find the shortest distances between words, but this got TLE. The alternative approach would be to just map the occurrences into a dictionary and return the return the minimum absolute difference between the two lists in linear time. This gets AC, but repeatedly calling this more than N times would in worst case have a higher complexity than O(n^2) time.

Complexity: O(n) time for init and shortest. O(n) space.

from collections import defaultdict
import sys

class WordDistance:

    def __init__(self, words):
        :type words: List[str]
        self.d = defaultdict(list)
        for i, word in enumerate(words):

    def shortest(self, word1, word2):
        :type word1: str
        :type word2: str
        :rtype: int

        if word1 == word2:
            return 0

        dist = sys.maxsize
        for loc1 in self.d[word1]:
            for loc2 in self.d[word2]:
                dist = min(dist, abs(loc1 - loc2))
        return dist

# Your WordDistance object will be instantiated and called as such:
# obj = WordDistance(words)
# param_1 = obj.shortest(word1,word2)

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