What is the time complexity of Design Add and Search Words Data Structure?

The optimal solution has a time complexity of O(m*26^n).

What is the space complexity of Design Add and Search Words Data Structure?

The space complexity is O(m*n).

What data structures are used to solve Design Add and Search Words Data Structure?

This problem uses String, Depth-First Search, Design, Trie. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Design Add and Search Words Data Structure?

Insight 1: A Trie (prefix tree) is a suitable data structure for storing the words, as it allows efficient prefix-based searching. Insight 2: The wildcard character '.' in the search operation necessitates a depth-first search (DFS) approach to explore all possible matching paths in the Trie.

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Design Add and Search Words Data Structure - LeetCode 211 Solution

Q: Which companies ask LeetCode 211. Design Add and Search Words Data Structure in interviews?

This problem is frequently asked at Atlassian, Datadog, Docusign, DoorDash, LinkedIn and 2 more companies.

Design Add and Search Words Data Structure - Complete Solution Guide

Design Add and Search Words Data Structure is LeetCode problem 211, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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. void addWord(word) Adds word to the data structure, it can be matched later. bool search(word) Returns true if there is any string in the data structure that matches word or false otherwise. word may contain dots '.' where dots can be matched with any letter. Example: Input ["WordDictionary","addWord","add

Detailed Explanation

The problem requires designing a data structure, `WordDictionary`, that supports two operations: `addWord(word)` and `search(word)`. `addWord` stores a given word into the data structure. `search(word)` checks if a given word (which may contain the wildcard character '.') exists in the data structure. The '.' character can match any single character. Essentially, we need to implement a dictionary that supports regular expression-like searching where '.' acts as a single-character wildcard.

Solution Approach

The solution uses a Trie data structure to store the words added to the dictionary. The `addWord` function inserts the word into the Trie, creating new nodes for characters that don't exist. The `search` function performs a depth-first search (DFS) on the Trie. If the current character in the search word is a '.', it explores all possible children of the current Trie node. If it's a normal character, it checks if that character exists as a child and recursively continues the search. The DFS returns true if a matching path is found, and false otherwise.

Step-by-Step Algorithm

Step 1: **Initialize Trie:** The `WordDictionary` is initialized with an empty Trie, typically represented as a dictionary or hash map.
Step 2: **`addWord(word)`:** Iterate through the characters of the input `word`. For each character, check if it exists as a child node in the current Trie node. If not, create a new child node for that character. Move to the child node. After processing all characters, mark the last node as the end of a word by adding a special character like '$' to the node's children.
Step 3: **`search(word)`:** Implement a depth-first search (DFS) function. The DFS function takes the current Trie node, the search word, and the current index in the search word as input.
Step 4: **DFS Base Case:** If the index reaches the end of the search word, check if the current Trie node is marked as the end of a word (has the '$' child). If so, return true; otherwise, return false.
Step 5: **DFS Wildcard Handling:** If the current character in the search word is '.', iterate through all the children of the current Trie node (excluding the end-of-word marker). Recursively call the DFS function for each child node with the index incremented.
Step 6: **DFS Character Matching:** If the current character in the search word is not a '.', check if it exists as a child in the current Trie node. If not, return false. If it exists, recursively call the DFS function with the child node and the index incremented.
Step 7: **DFS Return:** The `search` function initiates the DFS from the root of the Trie and returns the result of the DFS.

Key Insights

Insight 1: A Trie (prefix tree) is a suitable data structure for storing the words, as it allows efficient prefix-based searching.
Insight 2: The wildcard character '.' in the search operation necessitates a depth-first search (DFS) approach to explore all possible matching paths in the Trie.
Insight 3: The end-of-word marker is crucial for distinguishing between prefixes and complete words during the search.

Complexity Analysis

Time Complexity: O(m*26^n)

Space Complexity: O(m*n)

Topics

This problem involves: String, Depth-First Search, Design, Trie.

Companies

Asked at: Atlassian, Datadog, Docusign, DoorDash, LinkedIn, Rubrik, Snowflake.

Design Add and Search Words Data Structure | LeetCode 211

Medium

String Depth-First Search Design Trie

Time: O(m*26^n)Space: O(m*n)

Solution Code

Available in:

import java.util.HashMap;
import java.util.Map;

class WordDictionary {

    private Map<Character, Object> trie;

    public WordDictionary() {
        trie = new HashMap<>();
    }

    public void addWord(String word) {
        Map<Character, Object> node = trie;
        for (char charr : word.toCharArray()) {
            node = (Map<Character, Object>) ((Map) node).computeIfAbsent(charr, k -> new HashMap<>());
        }
        ((Map) node).put('$', true);
    }

    public boolean search(String word) {

        return dfs(trie, word, 0);
    }

    private boolean dfs(Map<Character, Object> node, String word, int index) {
        if (index == word.length()) {
            return node.containsKey('$');
        }

        char charr = word.charAt(index);

        if (charr == '.') {
            for (Character key : node.keySet()) {
                if (key != '$') {
                    if (dfs((Map<Character, Object>) node.get(key), word, index + 1)) {
                        return true;
                    }
                }
            }
            return false;
        } else {
            if (!node.containsKey(charr)) {
                return false;
            }
            return dfs((Map<Character, Object>) node.get(charr), word, index + 1);
        }
    }
}

/**
 * Your WordDictionary object will be instantiated and called as such:
 * WordDictionary obj = new WordDictionary();
 * obj.addWord(word);
 * boolean param_2 = obj.search(word);
 */

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Problem Info

Description

Problem 211: Design Add and Search Words Data Structure - Detailed explanation and solution approaches.

Topics

String Depth-First Search Design Trie

Companies

AtlassianDatadogDocusignDoorDashLinkedInRubrikSnowflake

Stats

Difficulty:Medium

Languages:4

Companies:7

Problem #:211

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

7 steps

1
Step 1: **Initialize Trie:** The `WordDictionary` is initialized with an empty Trie, typically represented as a dictionary or hash map.
2
Step 2: **`addWord(word)`:** Iterate through the characters of the input `word`. For each character, check if it exists as a child node in the current Trie node. If not, create a new child node for that character. Move to the child node. After processing all characters, mark the last node as the end of a word by adding a special character like '$' to the node's children.
3
Step 3: **`search(word)`:** Implement a depth-first search (DFS) function. The DFS function takes the current Trie node, the search word, and the current index in the search word as input.
4
Step 4: **DFS Base Case:** If the index reaches the end of the search word, check if the current Trie node is marked as the end of a word (has the '$' child). If so, return true; otherwise, return false.
5
Step 5: **DFS Wildcard Handling:** If the current character in the search word is '.', iterate through all the children of the current Trie node (excluding the end-of-word marker). Recursively call the DFS function for each child node with the index incremented.
6
Step 6: **DFS Character Matching:** If the current character in the search word is not a '.', check if it exists as a child in the current Trie node. If not, return false. If it exists, recursively call the DFS function with the child node and the index incremented.
7
Step 7: **DFS Return:** The `search` function initiates the DFS from the root of the Trie and returns the result of the DFS.

💡

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*26^n)

💾

Space Complexity

O(m*n)

📝

Why?

The time complexity for `search` is exponential because, in the worst case with all '.' characters, the algorithm explores every possible character at each level of the search. The space complexity is determined by the number of nodes created in the Trie, which depends on the number and length of the added words.

📖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

Using a List or Set of Words

Instead of using a Trie, one could store the words in a list or a set. The `search` function would then iterate through the list/set and compare each word with the input word, handling the '.' character with regular expressions or string comparison logic.

Time: addWord: O(1) for list, O(log n) for set. search: O(n * k) where n is the number of words and k is the average length of the words. The search needs to compare each word in the set/listSpace: O(n * k), where n is number of words and k is the average length

Trade-offs: Using a list/set is simpler to implement but less efficient for searching, especially with wildcards. The Trie is much more performant when dealing with numerous searches and words with common prefixes. If the frequency of `addWord` is much higher than `search`, this approach could be considered, however Trie is generally preferred

✨

Key Insights

1Insight 1: A Trie (prefix tree) is a suitable data structure for storing the words, as it allows efficient prefix-based searching.
2Insight 2: The wildcard character '.' in the search operation necessitates a depth-first search (DFS) approach to explore all possible matching paths in the Trie.
3Insight 3: The end-of-word marker is crucial for distinguishing between prefixes and complete words during the search.

📚

Key Terms Explained

For Students

String

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

Edge Cases

Edge case 1: Empty Trie. The `search` function should return false if the Trie is empty.
Edge case 2: Empty search word. The search word is an empty string. Should return true if the root node is marked as the end of a word; otherwise, return false.
Edge case 3: Word with multiple consecutive dots, e.g., 'b..'. The algorithm correctly handles such cases by exploring all possible paths in the Trie for each dot.
Edge case 4: Adding duplicate words. The Trie implementation efficiently handles this as inserting an existing word doesn't create new nodes, only marks the final node as end-of-word if it wasn't already.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting the end-of-word marker. Failing to mark the end of a word in the Trie can lead to incorrect search results, as prefixes of added words might be incorrectly identified as complete words.

✗Common mistake 2: Incorrectly handling the '.' character. Not exploring all possible paths in the Trie when encountering a '.' in the search word can result in missed matches.

✗Common mistake 3: Not considering the base case for DFS. Incorrectly handling the case where the search word is exhausted can lead to a stack overflow or incorrect search results.

✗Common mistake 4: Not freeing allocated memory (especially in C). This can cause memory leaks and degrade performance over time.

✅

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: Spell checking and auto-completion. A Trie-based dictionary can be used to suggest corrections for misspelled words and provide auto-completion suggestions as the user types.

🏦Real-world scenario 2: Implementing regular expression matching. The core logic of the `search` function can be adapted to support more complex regular expression patterns.

🎮Real-world scenario 3: Network routing. Tries can be used to store and efficiently search for IP addresses and network prefixes.

📱Real-world scenario 4: Bioinformatics. Tries can be used to store and search for DNA sequences.

💡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

String problems often involve sliding windows or two pointers. Consider if the character set is limited (lowercase letters = only 26 possibilities).

🎯 Interview Tip

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

Mutating strings directly in languages where strings are immutable. Use StringBuilder or arrays.

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: July 2, 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

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 13

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

@InterviewProJul 6

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

@CodeNinjaJun 29

❓

Frequently Asked Questions

What is the sliding window technique for string problems?

Sliding window maintains a "window" of characters with two pointers. Expand the right pointer to include characters, shrink from left when window violates constraints. Useful for substring problems, anagrams, and longest/shortest substring with properties.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Design Add and Search Words Data Structure"?

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

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