Leetcode: Shortest Word Distance II

Question

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.

Hide Tags Hash Table Design
Hide Similar Problems

–

Solution

Get idea from here.

class WordDistance(object):    def __init__(self, words):        """        initialize your data structure here.        :type words: List[str]        """        self.wordslen = len(words)        self.dictn = {}        for ind,elem in enumerate(words):            if elem in self.dictn:                self.dictn[elem] += [ind]            else:                self.dictn[elem] = [ind]    def shortest(self, word1, word2):        """        Adds a word into the data structure.        :type word1: str        :type word2: str        :rtype: int        """        self.minv = self.wordslen        l1 = self.dictn[word1]        l2 = self.dictn[word2]        ind1, ind2 = 0, 0        while ind1<len(l1) and ind2<len(l2):            a, b = l1[ind1], l2[ind2]            if a<b:                self.minv = min(self.minv, b-a)                ind1 += 1            else:                self.minv = min(self.minv, a-b)                ind2 += 1        return self.minv# Your WordDistance object will be instantiated and called as such:# wordDistance = WordDistance(words)# wordDistance.shortest("word1", "word2")# wordDistance.shortest("anotherWord1", "anotherWord2")