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Tested Python solution for LeetCode 243 with 15 pytest cases. Generate a practice environment with lcpy.
lcpy gen -n 243 # by problem number
lcpy gen -s shortest_word_distance # by problem name
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.
Input: wordsDict = ["practice", "makes", "perfect", "coding", "makes"], word1 = "coding", word2 = "practice"
Output: 3
Input: wordsDict = ["practice", "makes", "perfect", "coding", "makes"], word1 = "makes", word2 = "coding"
Output: 1
2 <= wordsDict.length <= 3 * 10^41 <= wordsDict[i].length <= 10wordsDict[i] consists of lowercase English letters.word1 and word2 are in wordsDict.word1 != word2class Solution:
# Time: O(n)
# Space: O(1)
def shortest_distance(self, words_dict: list[str], word1: str, word2: str) -> int:
index1 = -1
index2 = -1
shortest = len(words_dict)
for i, word in enumerate(words_dict):
if word == word1:
index1 = i
if word == word2:
index2 = i
if index1 != -1 and index2 != -1:
shortest = min(shortest, abs(index1 - index2))
return shortest
| Time | Space |
|---|---|
| O(n) | O(1) |