Leetcode #211: Design Add and Search Words Data Structure
In this guide, we solve Leetcode #211 Design Add and Search Words Data Structure 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 supports adding new words and finding if a string matches any previously added string. Implement the WordDictionary class: WordDictionary() Initializes the object.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Depth-First Search, Design, Trie, String
Intuition
We need to explore a structure deeply before backing up, which suits DFS.
DFS keeps local context on the call stack and is easy to implement recursively.
Approach
Define a recursive DFS that carries the necessary state.
Combine child results as the recursion unwinds.
Steps:
- Define a recursive DFS with state.
- Visit children and combine results.
- Return the final aggregation.
Example
Input
["WordDictionary","addWord","addWord","addWord","search","search","search","search"]
[[],["bad"],["dad"],["mad"],["pad"],["bad"],[".ad"],["b.."]]
Output
[null,null,null,null,false,true,true,true]
Explanation
WordDictionary wordDictionary = new WordDictionary();
wordDictionary.addWord("bad");
wordDictionary.addWord("dad");
wordDictionary.addWord("mad");
wordDictionary.search("pad"); // return False
wordDictionary.search("bad"); // return True
wordDictionary.search(".ad"); // return True
wordDictionary.search("b.."); // return True
Python Solution
class Trie:
def __init__(self):
self.children = [None] * 26
self.is_end = False
class WordDictionary:
def __init__(self):
self.trie = Trie()
def addWord(self, word: str) -> None:
node = self.trie
for c in word:
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, word: str) -> bool:
def search(word, node):
for i in range(len(word)):
c = word[i]
idx = ord(c) - ord('a')
if c != '.' and node.children[idx] is None:
return False
if c == '.':
for child in node.children:
if child is not None and search(word[i + 1 :], child):
return True
return False
node = node.children[idx]
return node.is_end
return search(word, self.trie)
# Your WordDictionary object will be instantiated and called as such:
# obj = WordDictionary()
# obj.addWord(word)
# param_2 = obj.search(word)
Complexity
The time complexity is O(V+E). The space complexity is O(V).
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.