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Tested Python solution for LeetCode 245 with 18 pytest cases. Generate a practice environment with lcpy.
lcpy gen -n 245 # by problem number
lcpy gen -s shortest_word_distance_iii # by problem name
wordsDict and two strings that already exist in the array word1 and word2, return the shortest distance between the occurrence of these two words in the list.
Note that word1 and word2 may be the same. It is guaranteed that they represent two individual words in the list.
Input: wordsDict = ["practice", "makes", "perfect", "coding", "makes"], word1 = "makes", word2 = "coding"
Output: 1
Input: wordsDict = ["practice", "makes", "perfect", "coding", "makes"], word1 = "makes", word2 = "makes"
Output: 3
1 <= wordsDict.length <= 10^51 <= wordsDict[i].length <= 10wordsDict[i] consists of lowercase English letters.word1 and word2 are in wordsDict.class Solution:
# Time: O(n)
# Space: O(1)
def shortest_word_distance(self, words_dict: list[str], word1: str, word2: str) -> int:
n = len(words_dict)
ans = n
if word1 == word2:
prev = -1
for i, w in enumerate(words_dict):
if w == word1:
if prev != -1:
ans = min(ans, i - prev)
prev = i
else:
i = j = -1
for k, w in enumerate(words_dict):
if w == word1:
i = k
if w == word2:
j = k
if i != -1 and j != -1:
ans = min(ans, abs(i - j))
return ans
| Time | Space |
|---|---|
| O(n) | O(1) |