Leetcode #245: Shortest Word Distance III
In this guide, we solve Leetcode #245 Shortest Word Distance III in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
Given an array of strings 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.
Quick Facts
- Difficulty: Medium
- Premium: Yes
- Tags: Array, String
Intuition
We need to scan characters while tracking positions or counts.
A simple state machine keeps the logic precise.
Approach
Iterate through the string once and update the state for each character.
Use a map or array if you need fast lookups.
Steps:
- Iterate through characters.
- Maintain necessary state.
- Build or validate the output.
Example
Input: wordsDict = ["practice", "makes", "perfect", "coding", "makes"], word1 = "makes", word2 = "coding"
Output: 1
Python Solution
class Solution:
def shortestWordDistance(self, wordsDict: List[str], word1: str, word2: str) -> int:
ans = len(wordsDict)
if word1 == word2:
j = -1
for i, w in enumerate(wordsDict):
if w == word1:
if j != -1:
ans = min(ans, i - j)
j = i
else:
i = j = -1
for k, w in enumerate(wordsDict):
if w == word1:
i = k
if w == word2:
j = k
if i != -1 and j != -1:
ans = min(ans, abs(i - j))
return ans
Complexity
The time complexity is , where is the length of the array . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.