LeetCode #211 — MEDIUM

Design Add and Search Words Data Structure

Move from brute-force thinking to an efficient approach using design strategy.

Solve on LeetCode

The Problem

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","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

Constraints:

1 <= word.length <= 25
word in addWord consists of lowercase English letters.
word in search consist of '.' or lowercase English letters.
There will be at most 2 dots in word for search queries.
At most 10⁴ calls will be made to addWord and search.

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Design · Trie

Example 1

["WordDictionary","addWord","addWord","addWord","search","search","search","search"]
[[],["bad"],["dad"],["mad"],["pad"],["bad"],[".ad"],["b.."]]

Core Insight

What unlocks the optimal approach

You should be familiar with how a Trie works. If not, please work on this problem: <a href="https://leetcode.com/problems/implement-trie-prefix-tree/">Implement Trie (Prefix Tree)</a> first.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #211: Design Add and Search Words Data Structure
class Trie {
    Trie[] children = new Trie[26];
    boolean isEnd;
}

class WordDictionary {
    private Trie trie;

    /** Initialize your data structure here. */
    public WordDictionary() {
        trie = new Trie();
    }

    public void addWord(String word) {
        Trie node = trie;
        for (char c : word.toCharArray()) {
            int idx = c - 'a';
            if (node.children[idx] == null) {
                node.children[idx] = new Trie();
            }
            node = node.children[idx];
        }
        node.isEnd = true;
    }

    public boolean search(String word) {
        return search(word, trie);
    }

    private boolean search(String word, Trie node) {
        for (int i = 0; i < word.length(); ++i) {
            char c = word.charAt(i);
            int idx = c - 'a';
            if (c != '.' && node.children[idx] == null) {
                return false;
            }
            if (c == '.') {
                for (Trie child : node.children) {
                    if (child != null && search(word.substring(i + 1), child)) {
                        return true;
                    }
                }
                return false;
            }
            node = node.children[idx];
        }
        return node.isEnd;
    }
}

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

// Accepted solution for LeetCode #211: Design Add and Search Words Data Structure
type WordDictionary struct {
	root *trie
}

func Constructor() WordDictionary {
	return WordDictionary{new(trie)}
}

func (this *WordDictionary) AddWord(word string) {
	this.root.insert(word)
}

func (this *WordDictionary) Search(word string) bool {
	n := len(word)

	var dfs func(int, *trie) bool
	dfs = func(i int, cur *trie) bool {
		if i == n {
			return cur.isEnd
		}
		c := word[i]
		if c != '.' {
			child := cur.children[c-'a']
			if child != nil && dfs(i+1, child) {
				return true
			}
		} else {
			for _, child := range cur.children {
				if child != nil && dfs(i+1, child) {
					return true
				}
			}
		}
		return false
	}

	return dfs(0, this.root)
}

type trie struct {
	children [26]*trie
	isEnd    bool
}

func (t *trie) insert(word string) {
	cur := t
	for _, c := range word {
		c -= 'a'
		if cur.children[c] == nil {
			cur.children[c] = new(trie)
		}
		cur = cur.children[c]
	}
	cur.isEnd = true
}

/**
 * Your WordDictionary object will be instantiated and called as such:
 * obj := Constructor();
 * obj.AddWord(word);
 * param_2 := obj.Search(word);
 */

# Accepted solution for LeetCode #211: Design Add and Search Words Data Structure
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)

// Accepted solution for LeetCode #211: Design Add and Search Words Data Structure
#[derive(Default)]
struct WordDictionary {
    is_word_end: bool,
    children: [Option<Box<WordDictionary>>; 26],
}

impl WordDictionary {

    fn new() -> Self {
        Default::default()
    }
    
    fn add_word(&mut self, word: String) {
        let mut dict = self;
        
        for c in word.chars(){
            dict = dict.children[char_index(c)]
                .get_or_insert(Box::new(WordDictionary::new()))
                .as_mut();
        }
        
        dict.is_word_end = true;
    }
    
    fn search_ref(&self, word: &str) -> bool{
        let mut dict = self;
        
        for (i, c) in word.chars().enumerate(){
            if c == '.'{
                let res = dict.children
                    .iter()
                    .any(|opt|{
                        if let Some(ref next_dict) = opt{
                            next_dict.search_ref(&word[i + 1..])
                        }else{
                            false
                        }
                    });
                
                return res;
            }else{
                if let Some(ref next_dict) = dict.children[char_index(c)]{
                    dict = next_dict.as_ref();
                }else{
                    return false;
                }
            }
        }
        
        dict.is_word_end
    }
    
    fn search(&self, word: String) -> bool {
        self.search_ref(&word)
    }
}

pub fn char_index(c: char) -> usize{
    c as usize - 'a' as usize
}

// Accepted solution for LeetCode #211: Design Add and Search Words Data Structure
class TrieNode {
    children: { [key: string]: TrieNode };
    endOfWord: boolean;
    constructor() {
        this.children = {};
        this.endOfWord = false;
    }
}

class WordDictionary {
    root: TrieNode;
    constructor() {
        this.root = new TrieNode();
    }

    addWord(word: string): void {
        let cur = this.root;

        for (const c of word) {
            if (!(c in cur.children)) {
                cur.children[c] = new TrieNode();
            }
            cur = cur.children[c];
        }
        cur.endOfWord = true;
    }

    search(word: string): boolean {
        function dfs(j: number, root: TrieNode): boolean {
            let cur = root;

            for (let i = j; i < word.length; i++) {
                const c = word[i];

                if (c === '.') {
                    for (const key in cur.children) {
                        if (
                            Object.prototype.hasOwnProperty.call(
                                cur.children,
                                key,
                            )
                        ) {
                            const child = cur.children[key];
                            if (dfs(i + 1, child)) {
                                return true;
                            }
                        }
                    }
                    return false;
                } else {
                    if (!(c in cur.children)) {
                        return false;
                    }
                    cur = cur.children[c];
                }
            }

            return cur.endOfWord;
        }
        return dfs(0, this.root);
    }
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(1) per op

Space

O(n)

Approach Breakdown

NAIVE

O(n) per op time

O(n) space

Use a simple list or array for storage. Each operation (get, put, remove) requires a linear scan to find the target element — O(n) per operation. Space is O(n) to store the data. The linear search makes this impractical for frequent operations.

OPTIMIZED DESIGN

O(1) per op time

O(n) space

Design problems target O(1) amortized per operation by combining data structures (hash map + doubly-linked list for LRU, stack + min-tracking for MinStack). Space is always at least O(n) to store the data. The challenge is achieving constant-time operations through clever structure composition.

Shortcut: Combine two data structures to get O(1) for each operation type. Space is always O(n).

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.