Leetcode #244: Shortest Word Distance II
In this guide, we solve Leetcode #244 Shortest Word Distance II 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
Design a data structure that will be initialized with a string array, and then it should answer queries of the shortest distance between two different strings from the array. Implement the WordDistance class: WordDistance(String[] wordsDict) initializes the object with the strings array wordsDict.
Quick Facts
- Difficulty: Medium
- Premium: Yes
- Tags: Design, Array, Hash Table, Two Pointers, String
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
Input
["WordDistance", "shortest", "shortest"]
[[["practice", "makes", "perfect", "coding", "makes"]], ["coding", "practice"], ["makes", "coding"]]
Output
[null, 3, 1]
Explanation
WordDistance wordDistance = new WordDistance(["practice", "makes", "perfect", "coding", "makes"]);
wordDistance.shortest("coding", "practice"); // return 3
wordDistance.shortest("makes", "coding"); // return 1
Python Solution
class WordDistance:
def __init__(self, wordsDict: List[str]):
self.d = defaultdict(list)
for i, w in enumerate(wordsDict):
self.d[w].append(i)
def shortest(self, word1: str, word2: str) -> int:
a, b = self.d[word1], self.d[word2]
ans = inf
i = j = 0
while i < len(a) and j < len(b):
ans = min(ans, abs(a[i] - b[j]))
if a[i] <= b[j]:
i += 1
else:
j += 1
return ans
# Your WordDistance object will be instantiated and called as such:
# obj = WordDistance(wordsDict)
# param_1 = obj.shortest(word1,word2)
Complexity
The time complexity is O(n). The space complexity is O(n).
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.