243. Shortest Word Distance

243. Shortest Word Distance

Easy

Given an array of strings wordsDict and two different strings that already exist in the array word1 and word2, return the shortest distance between these two words in the list.

 

Example 1:

Input: wordsDict = ["practice", "makes", "perfect", "coding", "makes"], word1 = "coding", word2 = "practice"
Output: 3

Example 2:

Input: wordsDict = ["practice", "makes", "perfect", "coding", "makes"], word1 = "makes", word2 = "coding"
Output: 1

 

Constraints:

  • 2 <= wordsDict.length <= 3 * 104
  • 1 <= wordsDict[i].length <= 10
  • wordsDict[i] consists of lowercase English letters.
  • word1 and word2 are in wordsDict.
  • word1 != word2




 class Solution:
    def shortestDistance(self, wordsDict: List[str], word1: str, word2: str) -> int:

        w1 = -1
        w2 = -1
        res = sys.maxsize

        for i, word in enumerate(wordsDict):

            if word == word1:
                w1=i
            elif word == word2:
                w2=i

            if w1 != -1 and w2 != -1:
                res = min(res, abs(w2-w1))
                if res == 1:
                    return res

        return res

Random Note

From python 3.7 dict guarantees that order will be kept as they inserted, and popitem will use LIFO order but we need FIFO type system. so we need OrderedDict which have popIten(last = T/F) for this req. One thing, next(iter(dict)) will return the first key of the dict

Python