How do you solve Stream of Characters on LeetCode?

Stream of Characters (LeetCode #1032) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using a Trie allows for efficient suffix checking by leveraging character-based traversal.

What is the brute force solution for Stream of Characters?

The brute force approach for Stream of Characters is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks every possible suffix of the current stream against all words in the list. This is straightforward but inefficient, especially as the stream grows. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Stream of Characters?

Stream of Characters uses the following concepts: Array, String, Design, Trie, Data Stream. The recommended patterns to study are: Trie, Dynamic Programming.

What are the interview tips for Stream of Characters?

Clearly explain your thought process and why you choose a particular data structure. Discuss the trade-offs of different approaches, especially in terms of time and space complexity. Practice implementing Tries and understand their traversal methods.

What are common mistakes when solving Stream of Characters?

Not considering the need to reverse the words when building the Trie. Failing to manage the stream correctly, leading to incorrect suffix checks.

#1032

Stream of Characters

Hard

ArrayStringDesignTrieData StreamTrieDynamic Programming

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

Using a Trie allows us to efficiently check for suffixes by traversing the Trie in reverse. This is much faster than checking every suffix individually.

⚙️

Algorithm

4 steps

1Step 1: Build a Trie from the reversed words list.
2Step 2: For each character added to the stream, append it to the front of a stream string.
3Step 3: Traverse the Trie using the characters from the stream, checking if we reach a terminal node (indicating a match).
4Step 4: If we reach a terminal node during traversal, return true; otherwise, return false.

solution.py28 lines

1class TrieNode:
2    def __init__(self):
3        self.children = {}
4        self.is_end = False
5
6class StreamChecker:
7    def __init__(self, words):
8        self.trie = TrieNode()
9        for word in words:
10            node = self.trie
11            for char in reversed(word):
12                if char not in node.children:
13                    node.children[char] = TrieNode()
14                node = node.children[char]
15            node.is_end = True
16        self.stream = ''
17
18    def query(self, letter):
19        self.stream = letter + self.stream
20        node = self.trie
21        for char in self.stream:
22            if char in node.children:
23                node = node.children[char]
24                if node.is_end:
25                    return True
26            else:
27                break
28        return False

ℹ

Complexity note: The time complexity is O(n) because each query processes the character and traverses the Trie, which is efficient due to the structure of the Trie. The space complexity is O(n) due to storing the Trie nodes.

1Using a Trie allows for efficient suffix checking by leveraging character-based traversal.
2Reversing the words when inserting into the Trie enables direct suffix matching.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.