Leetcode #676: Implement Magic Dictionary
In this guide, we solve Leetcode #676 Implement Magic Dictionary 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 is initialized with a list of different words. Provided a string, you should determine if you can change exactly one character in this string to match any word in the data structure.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Depth-First Search, Design, Trie, Hash Table, 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
["MagicDictionary", "buildDict", "search", "search", "search", "search"]
[[], [["hello", "leetcode"]], ["hello"], ["hhllo"], ["hell"], ["leetcoded"]]
Output
[null, null, false, true, false, false]
Explanation
MagicDictionary magicDictionary = new MagicDictionary();
magicDictionary.buildDict(["hello", "leetcode"]);
magicDictionary.search("hello"); // return False
magicDictionary.search("hhllo"); // We can change the second 'h' to 'e' to match "hello" so we return True
magicDictionary.search("hell"); // return False
magicDictionary.search("leetcoded"); // return False
Python Solution
class Trie:
__slots__ = "children", "is_end"
def __init__(self):
self.children: List[Optional[Trie]] = [None] * 26
self.is_end = False
def insert(self, w: str) -> None:
node = self
for c in w:
idx = ord(c) - ord("a")
if node.children[idx] is None:
node.children[idx] = Trie()
node = node.children[idx]
node.is_end = True
def search(self, w: str) -> bool:
def dfs(i: int, node: Optional[Trie], diff: int) -> bool:
if i == len(w):
return diff == 1 and node.is_end
j = ord(w[i]) - ord("a")
if node.children[j] and dfs(i + 1, node.children[j], diff):
return True
return diff == 0 and any(
node.children[k] and dfs(i + 1, node.children[k], 1)
for k in range(26)
if k != j
)
return dfs(0, self, 0)
class MagicDictionary:
def __init__(self):
self.trie = Trie()
def buildDict(self, dictionary: List[str]) -> None:
for w in dictionary:
self.trie.insert(w)
def search(self, searchWord: str) -> bool:
return self.trie.search(searchWord)
# Your MagicDictionary object will be instantiated and called as such:
# obj = MagicDictionary()
# obj.buildDict(dictionary)
# param_2 = obj.search(searchWord)
Complexity
The time complexity is , and the space complexity is , where and are the number of words in the dictionary and the average length of the words, respectively, and is the number of words searched. The space complexity is , where and are the number of words in the dictionary and the average length of the words, respectively, and is the number of words searched.
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.