What is the time complexity of Stream of Characters?

The optimal solution has a time complexity of O(m).

What data structures are used to solve Stream of Characters?

This problem uses Array, String, Design, Trie, Data Stream. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Stream of Characters?

Insight 1: Using a Trie data structure allows for efficient prefix/suffix searching. Since we are checking for suffixes, we can build a Trie with the words reversed. This enables fast suffix lookup. Insight 2: Storing the incoming stream of characters in a queue (or deque) allows for quick access to the most recent characters, which form potential suffixes.

Which companies ask LeetCode 1032. Stream of Characters in interviews?

This problem is frequently asked at Jane Street, Salesforce.

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Stream of Characters - LeetCode 1032 Solution

Stream of Characters - Complete Solution Guide

Stream of Characters is LeetCode problem 1032, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Design an algorithm that accepts a stream of characters and checks if a suffix of these characters is a string of a given array of strings words . For example, if words = ["abc", "xyz"] and the stream added the four characters (one by one) 'a' , 'x' , 'y' , and 'z' , your algorithm should detect that the suffix "xyz" of the characters "axyz" matches "xyz" from words . Implement the StreamChecker class: StreamChecker(String[] words) Initializes the object with the strings array words . boolean qu

Detailed Explanation

The problem asks us to design a data structure that efficiently checks if a suffix of an incoming stream of characters matches any word in a given dictionary of words. The stream of characters arrives one character at a time, and after each character arrives, we need to determine if any suffix of the accumulated characters forms a word present in the dictionary. For example, if our dictionary contains ["abc", "xyz"] and the stream is 'a', 'x', 'y', 'z', then after receiving 'a' (stream is 'a') we return false, after 'x' (stream is 'ax') we return false, after 'y' (stream is 'axy') we return false, and after 'z' (stream is 'axyz') we return true because "xyz" is a suffix of "axyz" and is present in the dictionary.

Solution Approach

The solution utilizes a Trie data structure to store the reversed words from the input dictionary. A Trie is efficient for prefix-based searches, and reversing the words allows us to effectively search for suffixes. The solution also maintains a deque to store the recently seen characters from the stream. When a new character arrives, it is added to the deque. Then, the algorithm traverses the Trie, starting from the root and following the path dictated by the reversed deque contents. If a path exists in the Trie corresponding to a suffix of the deque and ends at a node marked as the end of a word, then the algorithm returns true. Otherwise, it returns false.

Step-by-Step Algorithm

Step 1: Initialize the Trie: Build a Trie data structure from the given list of words. Reverse each word before inserting it into the Trie. Mark the end of each reversed word in the Trie with a special flag (e.g., '#').
Step 2: Initialize the Character Stream History: Use a deque (double-ended queue) to keep track of the incoming characters from the stream. Set a maximum size for the deque equal to the length of the longest word in the dictionary.
Step 3: Process each query character: When a new character arrives, add it to the right end of the deque. If the deque's size exceeds the maximum length, remove the oldest character from the left end (to maintain a window of relevant suffixes).
Step 4: Search the Trie for a suffix match: Traverse the Trie, starting from the root, using the reversed characters in the deque. For each character, check if a corresponding branch exists in the Trie. If a character is not found in the Trie, it means no suffix starting with the current history can match, so return false.
Step 5: Check for end-of-word: If, during the Trie traversal, you reach a node that is marked as the end of a word (i.e., the '#' flag is set), it means that the current suffix matches a word in the dictionary, so return true.
Step 6: Return false if no match is found: If the Trie traversal completes without finding a matching suffix, return false.

Key Insights

Insight 1: Using a Trie data structure allows for efficient prefix/suffix searching. Since we are checking for suffixes, we can build a Trie with the words reversed. This enables fast suffix lookup.
Insight 2: Storing the incoming stream of characters in a queue (or deque) allows for quick access to the most recent characters, which form potential suffixes.
Insight 3: The maximum length of the stream history needed is the length of the longest word in the dictionary. Any characters before that cannot contribute to a valid suffix match.

Complexity Analysis

Time Complexity: O(m)

Space Complexity: O(S)

Topics

This problem involves: Array, String, Design, Trie, Data Stream.

Companies

Asked at: Jane Street, Salesforce.

Stream of Characters | LeetCode 1032

Hard

Array String Design Trie Data Stream

Time: O(m)Space: O(S)

Solution Code

Available in:

import java.util.ArrayList;
import java.util.Deque;
import java.util.HashMap;
import java.util.LinkedList;
import java.util.List;
import java.util.Map;

class StreamChecker {

    private Map<Character, Object> trie;
    private Deque<Character> history;
    private int maxWordLength;

    public StreamChecker(String[] words) {
        trie = new HashMap<>();
        maxWordLength = 0;
        for (String word : words) {
            if (word.length() > maxWordLength) {
                maxWordLength = word.length();
            }

            Map<Character, Object> node = trie;
            for (int i = word.length() - 1; i >= 0; i--) {
                char c = word.charAt(i);
                node = (Map<Character, Object>) node.computeIfAbsent(c, k -> new HashMap<Character, Object>());
            }
            ((Map<Character, Object>)node).put('#', true);
        }

        history = new LinkedList<>();
    }

    public boolean query(char letter) {
        history.addLast(letter);
        if (history.size() > maxWordLength) {
            history.removeFirst();
        }

        Map<Character, Object> node = trie;
        List<Character> reversedHistory = new ArrayList<>(history);
        java.util.Collections.reverse(reversedHistory);

        for (char c : reversedHistory) {
            if (node.containsKey(c)) {
                node = (Map<Character, Object>) node.get(c);
                if (((Map<Character, Object>)node).containsKey('#')) {
                    return true;
                }
            } else {
                break;
            }
        }

        return false;
    }
}

/**
 * Your StreamChecker object will be instantiated and called as such:
 * StreamChecker obj = new StreamChecker(words);
 * boolean param_1 = obj.query(letter);
 */

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Problem Info

Description

Problem 1032: Stream of Characters - Detailed explanation and solution approaches.

Topics

Array String Design Trie Data Stream

Companies

Jane StreetSalesforce

Stats

Difficulty:Hard

Languages:4

Companies:2

Problem #:1032

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

6 steps

1
Step 1: Initialize the Trie: Build a Trie data structure from the given list of words. Reverse each word before inserting it into the Trie. Mark the end of each reversed word in the Trie with a special flag (e.g., '#').
2
Step 2: Initialize the Character Stream History: Use a deque (double-ended queue) to keep track of the incoming characters from the stream. Set a maximum size for the deque equal to the length of the longest word in the dictionary.
3
Step 3: Process each query character: When a new character arrives, add it to the right end of the deque. If the deque's size exceeds the maximum length, remove the oldest character from the left end (to maintain a window of relevant suffixes).
4
Step 4: Search the Trie for a suffix match: Traverse the Trie, starting from the root, using the reversed characters in the deque. For each character, check if a corresponding branch exists in the Trie. If a character is not found in the Trie, it means no suffix starting with the current history can match, so return false.
5
Step 5: Check for end-of-word: If, during the Trie traversal, you reach a node that is marked as the end of a word (i.e., the '#' flag is set), it means that the current suffix matches a word in the dictionary, so return true.
6
Step 6: Return false if no match is found: If the Trie traversal completes without finding a matching suffix, return false.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(m)

💾

Space Complexity

O(S)

📝

Why?

The Trie's space complexity is proportional to the total number of characters in the words because each character corresponds to a node or a branch in the Trie. The time complexity of query is O(m) because, in the worst case, we iterate through all 'm' characters stored in the deque (history) to traverse the Trie.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Brute-Force Suffix Checking

For each incoming character, iterate through all the words in the dictionary and check if any suffix of the current stream matches the word. This would involve repeatedly extracting suffixes from the current stream and comparing them against all words in the dictionary.

Time: O(n*m*k), where n is the number of words, m is the maximum word length, and k is the current length of the character stream. In the worst case, where k can be large(4*10^4), the performance would be extremely slow.Space: O(k), where k is the current length of the character stream.

Trade-offs: This approach is simple to understand, but highly inefficient for large dictionaries and long streams of characters. It is practical only for very small input sizes.

✨

Key Insights

1Insight 1: Using a Trie data structure allows for efficient prefix/suffix searching. Since we are checking for suffixes, we can build a Trie with the words reversed. This enables fast suffix lookup.
2Insight 2: Storing the incoming stream of characters in a queue (or deque) allows for quick access to the most recent characters, which form potential suffixes.
3Insight 3: The maximum length of the stream history needed is the length of the longest word in the dictionary. Any characters before that cannot contribute to a valid suffix match.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Edge case 1: Empty word list: If the input word list is empty, the Trie will be empty. The query function will always return false, which is the correct behavior.
Edge case 2: Single character words: If the word list contains single character words, the algorithm correctly identifies them when they appear in the stream.
Edge case 3: Long words: The algorithm can handle arbitrarily long words in the word list, within the constraints of the maximum number of queries and word lengths provided in the problem description. The history deque size is limited to the maximum word length, preventing unbounded memory usage.
Edge case 4: Repeated characters in stream or words: The Trie handles repeated characters without issues, and the algorithm correctly identifies matches even with repeated characters.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not reversing the words when building the Trie. This makes it difficult to efficiently search for suffixes.

✗Common mistake 2: Not limiting the size of the character stream history. This can lead to excessive memory usage as the stream grows.

✗Common mistake 3: Forgetting to mark the end of words in the Trie. Without this, the algorithm may incorrectly identify prefixes as valid words.

✗Common mistake 4: Failing to handle the case where the input word list is empty.

✗Common mistake 5: Implementing the Trie node as a fixed-size array for children, which limits the characters that can be used (the problem statement says lowercase letters, but the array would need resizing if you want to use other characters)

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Autocomplete suggestions in search engines. As the user types, the search engine can use a Trie to quickly find words or phrases that start with the entered text (or end with, if the Trie stores reversed words for suffix-based completion).

🏦Real-world scenario 2: Network intrusion detection systems. The system can maintain a dictionary of known malicious patterns (e.g., URLs, byte sequences). As network traffic is analyzed, the system can use a similar approach to quickly identify if any of the patterns are present in the traffic stream.

🎮Real-world scenario 3: Spam filtering. Email systems can use a Trie containing known spam keywords or phrases to identify and filter out spam messages. This problem's solution applies well when analyzing email streams on the fly.

📱Real-world scenario 4: Code editors syntax highlighting: When coding, some libraries and compilers use tries to match the last characters and identify syntax elements to highlight dynamically.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: October 11, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 14

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 7

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How do I handle string manipulation efficiently?

In languages with immutable strings (Java, Python, C#), use StringBuilder/list to avoid O(n) concatenation cost. Consider using character arrays for in-place modifications. For pattern matching, consider KMP or Rabin-Karp for O(n+m) complexity.

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Stream of Characters"?

For this hard-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether string techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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