Leetcode #1804: Implement Trie II (Prefix Tree)
In this guide, we solve Leetcode #1804 Implement Trie II (Prefix Tree) 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
A trie (pronounced as "try") or prefix tree is a tree data structure used to efficiently store and retrieve keys in a dataset of strings. There are various applications of this data structure, such as autocomplete and spellchecker.
Quick Facts
- Difficulty: Medium
- Premium: Yes
- Tags: 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
["Trie", "insert", "insert", "countWordsEqualTo", "countWordsStartingWith", "erase", "countWordsEqualTo", "countWordsStartingWith", "erase", "countWordsStartingWith"]
[[], ["apple"], ["apple"], ["apple"], ["app"], ["apple"], ["apple"], ["app"], ["apple"], ["app"]]
Output
[null, null, null, 2, 2, null, 1, 1, null, 0]
Explanation
Trie trie = new Trie();
trie.insert("apple"); // Inserts "apple".
trie.insert("apple"); // Inserts another "apple".
trie.countWordsEqualTo("apple"); // There are two instances of "apple" so return 2.
trie.countWordsStartingWith("app"); // "app" is a prefix of "apple" so return 2.
trie.erase("apple"); // Erases one "apple".
trie.countWordsEqualTo("apple"); // Now there is only one instance of "apple" so return 1.
trie.countWordsStartingWith("app"); // return 1
trie.erase("apple"); // Erases "apple". Now the trie is empty.
trie.countWordsStartingWith("app"); // return 0
Python Solution
class Trie:
def __init__(self):
self.children = [None] * 26
self.v = self.pv = 0
def insert(self, word: str) -> None:
node = self
for c in word:
idx = ord(c) - ord('a')
if node.children[idx] is None:
node.children[idx] = Trie()
node = node.children[idx]
node.pv += 1
node.v += 1
def countWordsEqualTo(self, word: str) -> int:
node = self.search(word)
return 0 if node is None else node.v
def countWordsStartingWith(self, prefix: str) -> int:
node = self.search(prefix)
return 0 if node is None else node.pv
def erase(self, word: str) -> None:
node = self
for c in word:
idx = ord(c) - ord('a')
node = node.children[idx]
node.pv -= 1
node.v -= 1
def search(self, word):
node = self
for c in word:
idx = ord(c) - ord('a')
if node.children[idx] is None:
return None
node = node.children[idx]
return node
# Your Trie object will be instantiated and called as such:
# obj = Trie()
# obj.insert(word)
# param_2 = obj.countWordsEqualTo(word)
# param_3 = obj.countWordsStartingWith(prefix)
# obj.erase(word)
Complexity
The time complexity is , where is the length of the string. 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.